Molecular physics: laboratory practicum in physics 9786010421080

This manual describes the laboratory work in molecular physics at the rate of physical workshop. The manual is intended

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Molecular physics: laboratory practicum in physics
 9786010421080

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AL-FARABI KAZAKH NATIONAL UNIVERSITY

MOLECULAR PHYSICS LABORATORY PRACTICUM IN PHYSICS

Almaty «Qazaq university» 2016

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UDC 539.1 (076) M 79 Recommended for publication by the Scientific Council of the Faculty of Physics and Technology and RISO of Al-Farabi Kazakh National University (protocol No.1 from November 2, 2016 y.) Reviewers: Doctor of Physics and Mathematics sciences, Professor M.E. Abishev Doctor of technical sciences, Professor A.B. Ustimenko Authors: A.S. Askarova, S.S. Issatayev, S.A. Bolegenova, V.V. Kashkarov, I.N. Korzun, G. Toleuov, Sh.B. Gumarova, L.E. Strautman, O.A. Lavrichshev, M.S. Issatayev, Zh.K. Shortanbayeva

M 79

Molecular physics: laboratory practicum in physics / A.S. Askarova, S.S. Issatayev, S.A. Bolegenova, [et. al.]. – Almaty: Qazaq university, 2016. – 142 p. ISBN 978-601-04-2108-0 This manual describes the laboratory work in molecular physics at the rate of physical workshop. The manual is intended for undergraduate of the Faculty of Physics and Technology of Al-Farabi Kazakh National University. Published in authorial release. В практикуме даны руководства по выполнению лабораторных работ по молекулярной физике в условиях физической мастерской и их описания. Предназначен для студентов физико-технического факультета Казахского национального университета имени аль-Фараби. Издается в авторской редакции.

UDC 539.1 (076)

ISBN 978-601-04-2108-0

© Askarova A.S., Issatayev S.S., Bolegenova S.A., [et. al.], 2016 © Al-Farabi KazNU, 2016

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CONTENTS INTRODUCTION ..................................................................................... 4 Laboratory work No. 1 BINOMIAL LAW OF PROBABILITY DISTRIBUTION........................ 18 Laboratory work No. 2 CALIBRATION OF THERMOELECTRIC THERMOMETER ............... 32 Laboratory work No. 3 DETERMINATION OF BOLTZMANN CONSTANT ............................ 39 Laboratory work No. 4 DETERMINATION OF SPECIFIC HEAT OF TIN CRYSTALLIZATION (MELTING) AND CHANGES IN ENTROPY DURING CRYSTALLIZATION ...................................... 46 Laboratory work No. 5 DETERMINATION OF RELATION Ср/Сv USING THE STANDING WAVE METHOD ....................................................... 59 Laboratory work No. 6 DETERMINAITION OF SPECIFIC HEAT OF LIQUIDS BY COLORIMETRIC METHOD .............................................................. 68 Laboratory work No. 7 DETERMINATION OF AIR VISCOSITY ............................................. 77 Laboratory work No. 8 STUDY OF TEMPERATURE DEPENDENCE OF LIQUID VISCOSITY ............................................................................................. 95 Laboratory work No. 9 DETERMINATION OF THE SURFACE TENSION COEFFICIENT OF LIQUIDS .................................................................. 108 Laboratory work No. 10 DETERMINATION OF THERMAL CONDUCTIVITY OF SUBSTANCES BY COLORIMETRIC METHOD IN THE QUASI-STATIONARY REGIME .................................................... 117 APPENDIX ............................................................................................... 124

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INTRODUCTION The main objective of physical laboratory classes is quantitative study of physical phenomena, development of skills of independent research work and correct measurement of physical quantities. During laboratory classes students independently conduct physical experiments, get acquainted with the most important measuring instruments, acquire skills for carrying out precise measurements and correct processing of obtained results. Physical measurements During the experiments various types of measurements are made. Measurement is finding the value of a physical quantity empirically using special instruments. In the laboratory classes measurements are made to establish a functional relationship between two or more variables. There are several types of measurements. Their classification is based on the type of dependence of measured quantities on time, conditions determining the accuracy of measurements and methods of presentation of their results. According to the nature of dependence of measured quantities on the time of measurement, they are divided into static, when the measured value remains constant, for example, measurement of geometric dimensions of the body, constant pressure, and dynamic, when the measured value changes, for example, measurement of the amplitude of damped oscillations, pulsation rate, etc. According to the method of obtaining the results of measurements, they are divided into direct, indirect, cumulative and joint. Direct measurements are measurements where the sought value of a physical quantity is found directly using the measuring instrument. In this case, the value of the physical quantity is determined by direct contact with the measuring device, i.e. the quantitative value 4

of the unknown quantity is determined by the indications of the device, immediately, without further processing. For example, it is measuring of the size of the body using a micrometer, measuring time using stopwatch, etc. Direct measurements can be expressed by the formula x = Q, where x is the unknown quantity, and Q is the value directly derived from the experimental data. Direct measurements are widely used in mechanical engineering, industrial processes, and other spheres. Indirect measurements are measurements, in which the sought value is determined based on the known relationship between this value and the values subjected to direct measurements. Indirect measurements can be described by the formulas

x  f (Q1 , Q2 , Q3 ,...) , where х is the value of the unknown quantity; Q1 , Q2 , Q3 ,... are the values of quantities measured directly. Examples of indirect measurements are determination of the volume of the body from direct measurements of its geometric dimensions, finding of a specific electrical resistance of the conductor using its resistance, length and cross-section. Indirect measurements are widely used in cases when it is impossible or very difficult to find the unknown value as a result of direct measurements. Indirect measurements play an important role in measurements of quantities that cannot be measured by direct experimental comparison, for example, sizes of astronomical or intra-atomic level. Cumulative measurements are simultaneous measurements of several one-type quantities, in which the sought value is determined by solving the system of equations obtained by direct measurements of various combinations of these quantities. An example of cumulative measurements is determination of the mass of individual set of weights (calibration) by the known mass of one of them and by the results of direct comparisons of different combinations of weights. For example, it is necessary to calibrate a set of weights, consisting of weights 1, 2, 2*, 5, 10 and 20 kg (asterisk denotes the weight having the same nominal value). Calibration is determination of the mass of each weight using a reference weight, for example, weight of mass of 1 kg. To do this, 5

we make measurements, each time changing the combination of weights (numbers show masses of individual weights, 1ref means the mass of the reference weight of 1 kg) 1 = 1ref + a, 1 + 1ref = 2 + b, 2* = 2 + c, 1 + 2 + 2* = 5 + d , etc. Letters a, в, с and d denote weights that have to be added or subtracted from the weight indicated in the right-hand side of the equation. Solving this system of equations, we can determine the value of the mass of each weight. Joint measurements are simultaneous measurements of two or more different quantities aimed at finding dependencies between them. As an example, we can take simultaneous measurements of electrical resistance at 20 °C and temperature coefficients of the measuring resistor based on the direct measurements of its resistance at various temperatures. By the conditions determining the accuracy of the result, measurements are divided into three classes: 1. Measurements with maximum possible accuracy attainable at the current level of measuring techniques. These include reference measurements and measurements of physical constants. 2. Control-and-calibration measurements, the error of which cannot exceed a predetermined value. These include measurements carried out by state supervision laboratories to monitor the observance of state standards. 3. Technical measurements in which the error of the result is determined by the characteristics of measuring instruments. These include measurements made in scientific research, production, and others. Units of measurement of physical quantities As physical quantities are in a functional relationship, it is impossible to establish their units arbitrary. There are seven units of physical quantities, called the basic units in SI, which are established arbitrarily. These are units of the following physical quantities: 6

length (meter), mass (kilogram), time (sec), temperature (Kelvin), current (Amps), intensity (candela), the amount of substance (mole). International System of Units includes two units additional to the basic units of measurement. They are units for plane (radians) and solid (steradian) angles. The units of other physical quantities are found from the relations determining these values. Therefore they are called derived units. Measurement errors It is impossible to measure the value of a physical quantity with absolute accuracy. Firstly, there is no such a precision instrument, and secondly, the possibilities of human senses are limited. Therefore, when it comes to the measured value, it is necessary to specify the error of measurement. By the nature of occurrence the errors are divided into systematic, random and gross errors. Systematic errors are values constant in time or varying regularly during the measuring experiment, for example, the initial deflection of the arrow of the device from zero. Systematic errors are subdivided into instrumental and methodical errors. Errors of measuring instruments Accuracy of measurement of any measuring instrument is limited. Errors of instruments intended for industrial purposes (ammeters, voltmeters, potentiometers, etc.) are determined by their measurement accuracy. Measurement accuracy of the instrument is usually expressed in percentage form. For example, an ammeter with an accuracy class of 0.2% determines the current value corresponding to the full-scale, with an error not exceeding 0.2%. The accuracy class of the device, regardless of the position of the arrow, indicates the minimum relative error, and relative error of the measured value increases as the arrow moves to the origin of coordinates. Therefore, during measurements we have to try to work in the second half of the scale. It is important at which angle we look at the arrow when you register the readings. To minimize the errors, we must look at the arrow of the device at the right angle, i.e. perpendicular to the scale. 7

Methodological errors Such errors are, for example, errors of measurement techniques. For example, it is not taking into account the friction of the shaft in the laboratory work No. 3 in determining the moment of inertia of the disc by dynamic method or using the real liquid (water) to check the Bernoulli equation obtained for an ideal fluid. Errors in the methods of data processing They arise as a result of incorrect or imperfect methodology used to determine the sought value by computing and determining the errors. As the number of measurements increases, systematic errors decrease. To reduce them, more precise tools and more advanced measurement and data processing techniques are used. Random errors Random errors are components of the measurement error, changing randomly in repeated measurements of the same quantity. These include not only random variations in the mains voltage and the temperature in the laboratory but also factors associated with the imperfection of human senses and skills. In the process of measuring, the values and signs of random errors do not remain constant. To reduce the random error, it is necessary: 1) to reduce the influence of external factors; 2) to measure carefully and thoroughly; 3) to increase the number of tests. Gross errors Gross errors are errors of measurement, significantly exceeding expectations under given conditions. They are caused by the negligence of the experimenter or by faulty equipment. When gross errors are found, such measurements are deleted. Methods of finding errors of direct measurements are described in detail in the laboratory work No. 1, so we do not consider them here, and continue with the method of finding errors of indirect measurements. 8

Finding errors of indirect measurements Let us consider some special cases of relations between two or more variables functionally connected with each other. 1. Let us consider the sum or difference of two quantities: A  BC.

(1)

The most probable values of quantities B and C are their arithmetic mean values  B  and  C  . Here and below the angular brackets denote the arithmetic mean values. The best value of A is equal to the sum (or difference) of the best values of terms: (2) A  B    C  . The arithmetic mean error S A , if В and С are independent, is found using the formula:

S A  S B2  S C2 ,

(3)

in other words, not the errors are added but their squares. 2. Let the unknown quantity be a product or a quotient of two other quantities: A=В·С or A=В/С.  (4) The most probable values of A are given by the formulas: or A  B  .

A  B    C 

C 

(5)

Relative standard errors of the product and quotient of independent variables are found using the formula: 2

2

SA  S   S    B   c  . A  B  C 

(6)

3. Let us consider the product of several variables having different powers: 9

A  B  C   E  ...

(7)

The relative standard error of the value of A for independent B, C, E,… can be obtained using the formula 2

2

2

2

 SA  2 S  2 S  2 S       B     C     E   ...  A B C E

(8)

4. Let us derive the general formula for calculations. Let

A  f ( B, C, E,...) ,

(9)

where f is an arbitrary function of В, С, Е etc. Then the best value of A is:

Aíàèë  f ( B ,  C ,  E ,...) .

(10)

The mean-square error of A is found as: 2

2

2

 f   f  2  f  2 S    S B2    S C    S E  ...  B   C   E  2 A

(11)

The notation f has the usual meaning of the partial derivative B

of function f with respect to B, i.e. the derivative to find which we assume all arguments except B constant. Partial derivatives should be calculated for the best values of arguments Вbest, Сbest, Еbest , etc. Let us consider some consequences that can be derived from the analysis of the above formulas. 1. One should avoid measurements in which the sought value is found as the difference of two large numbers. Thus, we should not dedtermine the thickness of the pipe wall by subtracting its inner diameter from the outer (and dividing the result by two). The relative error of measurement in this case is greatly increased as the measured value – in this case, the wall thickness is small, and the error in its definition is found by adding the errors of measurements of both diameters and therefore it increases. 10

2. One should determine all measured values with approximately the same relative error. For example, if the volume of the body is measured with an accuracy of 1%, there is no sense to spend time and efforts on measurement of the body weight with an accuracy of 0.01% in order to determine its density. 3. In calculating the error of the product of power functions it is necessary to pay attention to the member of the highest power. According to the formula (8), the relative error of the factor is directly proportional to the square of its power. Therefore, the errors of individual members can be neglected. In real experiments, systematic and random errors are present simultaneously. Typically, they are independent, so the total error can be determined by the following formula: S2tot = S2 ran +S2sys.

(12)

Approximate calculations and rounding rules In the laboratory work it is often necessary to measure several quantities with different accuracy. If the quantities in the formula have different errors, it is necessary to pay attention to the value of the maximum error. Then, when determining the total accuracy it is not necessary to calculate errors of quantities, the accuracy of which is many times smaller than the value of the maximum error. Sometimes, they are simply rejected. The accuracy of calculation of the required quantity must be within the measurement accuracy. When the accuracy of calculations is much higher than the measurement accuracy, it is considered a drawback of the work, rather than its advantage. Therefore, it is important to know the basic rules of rounding. Rounding rules and approximate calculations Rounding the number, we have to delete the last digits, and if the first of the deleted digits is less than 5, we do not change the remaining number, but if the first of the deleted digits is more than 5, we must add one to the previous number. If the first digit of the deleted part is equal to 5, there are two options. If the preceding digit is odd, then it is increased by one, if it is even – nothing is added. For example: a = 21.314  21.31;  = 3.141593  3.1416,  = 3.141593  3.14; l = 12,15 m  12,2 м; t = 5.25 s  5.2 s. 11

The value of the physical quantity carries information about the amount of error. In fact, if all of the numbers are correct, then the price of last digit gives the accuracy of the measurement or calculation. For example: t = 5.021000 s, in this record of time the error can be estimated approximately as 1 microsecond. Significant digits. Significant digits are all correct digits except zero, standing in front of the number. For example, the number 0.000205 has three significant digits. The first four zeros to the left are insignificant digits, zero between two and five is a significant digit. In the number 2300 there are four significant digits. Recording 6.1 ∙ 103 means that the number has only two significant digits (six and one). In the final recording the measurement error is usually expressed by the number with one or two significant digits. Two digits are used for more precise measurements, and if the higher digit of the number expressing error is three or less than three. For the least significant digit only 5 is used. It should be noted that in the intermediate calculations, in order the rounding errors not to distort the results too much, it is recommended to preserve three to four significant digits, depending on the computations. Some recommendations to the fulfillment of the laboratory work The laboratory work can be divided into four stages. 1. Preparatory stage. Each lab work is devoted to the study of a concrete physical phenomenon or process, the specific features of which are to be identified. On the first day of physical practicum all students are given the theme of the laboratory work, which they are to do at the next lesson. Before the lab work, they should at home (in the library) get familiar with the theory of this work, the construction of the experimental device (drawings) or the experimental model and the experimental part of laboratory work. The main source of information is a methodological description of the lab. From it, the student writes down the name and purpose of the work, a brief theory of the problem, describes the setup, the process of t work fulfillment and methods of mathematical processing of the results. To the laboratory work the student brings his drawing tools, graph paper, multi-functional calculator, etc. 12

2. Experimental part. This stage of physical practicum is fulfilled directly in the laboratory. The students who showed good preparation to the performance of the work are allowed to do it. The necessary work equipment, tools and details are given to the students by the laboratory assistant. The students make measurements without assistance. The students write table values of physical constants, parameters of the setup and the data of the experiment into the lab journal. At the end of the experiment the teacher checks the correctness of measurements, If everything is correct, then the teacher signs in the lab journal. After the signature of the teacher it is prohibited to make any changes in the data table. 3. Mathematical processing of experimental results. In this part of the laboratory work the sought value is calculated, graphs of the processes are plotted, measurement errors are calculated, and the results are analyzed. In the analysis of the results it is necessary to pay attention to the factors influencing the results of the experiments. 4. Defense of the results of laboratory work. In this final part the student justifies his results making an oral presentation to the teacher. He tells about the theory of the subject, derives the final formula, and answers the questions given at the end of the description. Graphic method of data processing In the experimental physics graphs are used for different purposes. For example:  to determine values;  to present visual description of the process;  to find an empirical relationship between the values;  to compare the experimental results with the theoretical data and the data of other authors, etc. Graphs are plotted on linear and logarithmic graph paper. To determine the functional dependence of the value it is accepted to put the value of the argument along the x-axis, and the values of the function – along the y-axis. Before plotting it is necessary to specify the minimum and maximum values for the abscissa and ordinate axes and to choose the most appropriate scale. The experimental results are shown on the graph as points and theoretical results – as a solid line. If several lines, corresponding to the different modes of the same process, are drawn on the same plot, it gives a visual representation 13

of the dynamics of the process, depending on its parameters. Optimally choosing the scale, it is necessary to find such a scale that the data points on the graph cover all the picture plane. Typically, the points on the graph are located in a specific pattern. If any point does not follow the dependence (lies aside), you should pay attention to it: either there is an interesting effect or a gross error. If the measurement is correct, then the vicinity of the point is investigated in more detail in order to find out the effect. By the size of the symbols on the graph, we can judge about the measurement errors. Dimensions of the graphic characters are defined by the expression: 1 l  S . 2 Here l is the size of linear characters. In the construction of graphs the students should try to draw a linear dependence as it is easy «to read» a linear functional dependence of variables and based on such a dependence it is not difficult to get an empirical relationship. Let us consider several ways of linearization of dependencies. 1. Let the process be described by a quadratic function. Consider the law of falling bodies described by the formula: S

gt 2 . 2

If we put S and t along the axes, the point will lie on the parabola, and it is almost impossible to draw it by eye. But if we put S and t2 or S and t or ln S and ln t along the axes, the graph will become a straight line. In general, transforming the quadratic dependence to the form y  ax  b , it is possible to determine the unknown constants a and b. 2. Let the process be described by a power dependence y  x n . Let us take logarithms of both sides of this formula:

ln y  n ln x . 14

If we put variables ln y and ln x along the axes, we will get a straight line. Least square method In general, many complex expressions can be reduced to the linear form: (13) y  ax  b . The parameters a and b in this relationship can be found analytically by using the experimental data. Let in the experiment we got the value y i ( i  1,2,..., n ), corresponding to x i . Then the values of a and b in formula (13) corresponding to the most precise dependence between y i and x i can be found by the least square method. If the experimental values x i and y i are substituted into (13), due to the errors in measurements the equation (13) is usually not valid, i.e. (14) yi  (axi  b)  0 . Let us square this difference. Then, find the sum of the squares of these differences for all measurements: n

2

D    yi  (axi  b) .

(15)

i 1

Using the least square method, we get minimum values of parameters a and b corresponding to the most accurate approximation of the linear dependence between y and x in the formula (15). Therefore, to find the values of parameters a and b, from the equation (15) we find partial derivatives of D with respect to a and b, and equate the resulting expressions to zero. n D  2  yi  (axi  b)  0 , b i 1

n D  2{ yi  (axi  b) xi }  0 . a i 1

15

(16) (17)

Then: n

y i 1

n

y x i 1

i i

n

 nb  a xi  0 ,

i

(18)

i 1

n

n

i 1

i 1

 b xi  a xi2  0 .

(19)

Solving the resulting system of equations, we can find the values of parameters a and b for the most accurate approximating linear dependence: n

a

n

i 1

tes:

i

i 1

 n xi yi

i

i 1

2

    xi   n   xi2 i 1  i 1  n

n

b

n

 y  x

,

n

n

n

i 1

i 1

 xi   xi yi   yi  xi2 i 1

i 1

(20)

n

2

.

(21)

n  n    xi   n   xi2 i 1  i 1 

If the linear dependence passes through the origin of coordinay  ax ,

(22)

then n

a

x y i 1 n

i

x i 1

i

.

(23)

2 i

The least square method allows us to find the errors of parameters in the linear dependence, which are determined by formulas (20) and (21): 2

2

n n  n   n    yi   n xi y i   xi  y i  i 1 i 1  , S 02  i 1   i 1    i 1 2 n  2 nn  2  n  n   nn  2n xi2    xi    i 1    i 1 n

 yi2

16

(24)

S a2 

nS 02   n xi2    xi  i 1  i 1  n

n

2

,

(25)

.

(26)

n

Sb2 

S 02  xi2 i 1

 n  n xi2    xi  i 1  i 1  n

2

Here S 0 is the mean square error of an individual measurement, S a and S b are the mean square errors of parameters in the linear dependence y  a x  b .

17

Laboratory work →1 BINOMIAL LAW OF PROBABILITY DISTRIBUTION 1.1. The purpose of work: experimental verification of the binomial distribution and its comparison with the Laplace-Gauss distribution-and Gaussian distribution.

1.2. A brief theoretical introduction 1.2.1. Deterministic and statistical laws. The form of manifestation of causal relations in the laws of nature are divided into two classes: deterministic and statistical. For example, using the laws of mechanics and knowing the present position of the planets of the solar system we can predict with a high degree of certainty their position in any given time in future including a very accurate prediction of solar and lunar eclipses. This is an example of deterministic laws. However, not all the phenomena of the macrocosm can be precisely predicted, despite the fact that our knowledge is continuously deepened and refined. For example, tossing a coin, on the basis of the laws of mechanics it is practically impossible to accurately predict how it will fall: heads or tails. We can only speak with a certain probability about the result of tossing. Such laws are called statistical. Knowing that the probability of the eagle and tails is the same, we can state that throwing again the coin, we will get the eagle with probability 0.5, or 50%. This means that if we toss a coin 10 times in five cases we will get the eagle, and in the other five cases – tails. The more times we throw the coin, the more accurately this law will be fulfilled, we can verify experimentally statistical laws only after a large number of tests. In the example described, there is a principal opportunity to predict the result of a particular test if we exactly know how fast the coin was tossed, the angle of tossing, the resistance of the medium in which it moves, the acceleration of gravity at this point of the Earth, etc. i.e. all the factors influencing its movement. In fact, it is practi18

cally impossible to take into account the effect of absolutely all forces and influences, which superimposing make the result of tossing a random value. Even less deterministic are many laws of the microworld. For example, from the point of view of quantum physics we cannot speak about the position in the electron in a definite point in space and at a certain moment of time, we can only determine the state of the system with a certain probability, which is a measure of possibility of realization of previous tendencies. Molecular Physics occupies an intermediate position between the macroscopic processes obeying deterministic laws of nature, and quantum-mechanical phenomena, mainly described by statistical laws. Therefore processes occurring in gases and liquids, can be described both from the phenomenological point of view (thermodynamics) and using a statistical approach (the molecular-kinetic theory). For example, the binomial probability distribution is used in molecular physics to calculate the probability of the macrostate of the system, consisting of a large number of particles, as well as to determine the relationship between equilibrium and most probable states. In addition, based on the binomial distribution, we can obtain a mathematical relationship between the fluctuation and the number of particles in the system. In addition, there is a deep analogy between the model of an ideal gas and the Gaussian distribution of the probability density for continuous random variables. Using this distribution, it is possible to derive the Maxwell law for velocity and energy distributions. 1.2.2. Basic concepts of the probability theory. A certain event is an event that will obligatory happen if certain conditions are met. Hence, an impossible event is an event that will never happen under the given conditions. A random event is an event which for a given set of conditions can either happen or not happen. The measure of realization of such an event is probability. Elementary events are possible, mutually exclusive outcomes of the experiment. For example, tossing a single coin is an experiment, and an eagle (event A) and tails (event B) are elementary events. Thus, in this experiment, the number of elementary events is 2. 19

The probability P of an event A is the number of elementary events favorable for A, divided by the total number of all possible elementary events. If the experiment consists of a finite number of equiprobable elementary events, we have a classic case. Examples. 1. When the coin is tossed, the probability of an eagle is equal to the probability of tails, and is equal to 1/2. 2. When throwing the dice, each of the six faces of which have digits 1, 2, 3, 4, 5 and 6, the probability of falling of any of the numbers is the same and is equal to 1/6. 1.2.3. Binomial distribution. How to check that the probability of having, for example, an eagle is equal to 1/2? Obviously, it is necessary to conduct n experiments in which in 50% we have heads and in 50% – tails. In practice, for small n values, it may not be fulfilled, that is, for example, it is not necessary that out of 10 tosses 5 times the eagle will fall and 5 times – tails. In 100 tests this result will be fulfilled more precisely, in 1000 tests – even more precisely, etc. This distribution (50 x 50) will be exactly fulfilled for n   . However, in practice we can make only a finite number of trials. Therefore the question arises, what is the probability that out of n trials we get the eagle in n/2 trials? In a more general case we can determine the probability that in n experiments the heads falls 0,1,2,3, ..., k, ..., n times. Let the trial be repeated n times. The probability that in n trials the eagle will fall k times (i.e., event A occurs k times) is determined by the binomial distribution law:  k w ; p   C nk p k q n  k , n  

(1.1)

where Cnk  n! k!(n  k )! is the number of combinations from n with respect to k , p is the probability of eagle falling in a single tossing (or the probability of event A) q is the probability of not falling of an eagle in a single tossing (or the probability that in an individual experiment this event does not happen, that is the second event – tail falling – happens. 20

The formula (1.1) is derived in the APPENDIX 1. To simplify the task instead of tossing one coin n times throw n coins simultaneously. Then formula (1.1) can be interpreted as k follows: w ; p  is the probability that out of n tossed coins k coins n



will have eagles. However, to experimentally verify the formula (1.1), you must make the experiment with tossing n coins N times, and the more the N number is, the better agreement between the results of this experiment with the theoretical distribution (1.1) is obtained. The binomial distribution is valid if the following conditions are satisfied:  the number of tests n is fixed;  the outcome of each test does not depend on the outcome of the other tests (independent elementary events);  the probability p of the event A does not depend on the number of the test, i.e. p  const ;  the probability q that the event A will not happen is q  1  p . The combination of these conditions is called a mathematical model of the binomial experiment. Examples 1. The probability that out of 10 tosses of the coin eagle will fall 6 times is ( p  1 / 2): 6

w(6 / 10 ; 1 / 2) 

4

10 !  1   1  210  0,205. .     6 ! 4 !  2   2  1024

2. The probability that out of 3 tosses of the dice «6» will fall 3 times is ( p  1 / 6): 3

w (3 / 3 ; 1 / 6) 

0

3!  1   5  1  0,005.      3! 0!  6   6  216

3. The probability of the birth of a boy is p  0,515 . The probability that out of 10 newborn 5 will be boys is: w(5 / 10 ; 0,515 ) 

10 ! 0,515 5 0,485 5  252  0,036  0,027  0,245. 5! 5! 21

1.2.4. Properties of the binomial distribution. The quantity k in the formula (1.1) is a random, as we do not know exactly when this event will happen, and can predict it only with a certain probability. In the probability theory there are discrete and continuous random variables. A discrete random variable is a variable that takes a finite or countable number of values. This can be a set of numbers or a function taking discrete values. A continuous random variable is a variable taking a continuous set of values (remember the concept of continuity in the mathematical analysis). For example, the random variable k in formula (1.1) is discrete and the result of multiple measurements of the physical quantity a (which may be, for example, acceleration, characteristic dimensions of the body, thermal diffusivity, etc.) is a continuous random variable, as the measurements can give any values within a certain range. Mathematical expectation is defined as the arithmetic mean of the random variable found using the probability theory. Mathematically, it is defined by the formula: n k  mk   k  w  ; p . n  k 0

(1.2)

As shown in APPENDIX 2, for the binomial distribution (1.3)

mk  n p .

If p  q , the mathematical expectation coincides with the most probable value. It follows that the maximum probability of the experiment with the coin corresponds to the uniform distribution of coins (equal for two states). Variance is a mean value of squares of deviations of a random variable from its mathematical expectation, taking into account the probability:

 k2   k  mk 2 w  , p  . k n

n

k 0

22



(1.4)

Variance is a characteristic «spread» of values around its mean value. As shown in APPENDIX 2, for the binomial distribution

 k2  n p q . sics.

(1.5)

1.2.5. Application of the binomial distribution in molecular phy-

Probability of the macrostate Macrostate is a state consisting of a large number of microstates. If all the features characterizing the macrostate are known, it is possible, in principle, to list all microstates compatible with these features and calculate their number. Let Г  be the number of microstates, where α characterizes the macrostate, Ã 0 is the total number of states. Then, based on the postulate of equal probability of microstates, the probability of the considered macrostate is: w( )  Г Г 0 .

Let us introduce the following notations: – V is the volume occupied by the ideal gas; – n is the number of particles in the volume V; – N  V / d 3 is the number of cells that can be occupied by particles (d is the rib of the cell in the form of a cube). Here d3 10-30 m3, therefore N is very large, and the condition N  n is satisfied. Let us find the probability w (k / n; p) of the macrostate of the system, in which in a fixed volume V1 , forming part of the volume V, k particles out of the total number of n particles are present. According to the task V1  V , k  n . However, the volume V1 should not be too small to contain at least k cells in which k particles could be placed. The number of cells in the volume V1 is equal to N1  V1 / d 3 , therefore N1  k . It is assumed that particles are distinguishable from each other (e.g., numbered). This means that the two microstates, in which the 23

particles occupy the same cells, are different if, for example, two particles changed their places in some cells. However, the macroscopic state of the system from the rearrangement of particles in the cells will not change. We will show that this model corresponds to the mathematical model of the binomial experiment:  The number of particles in the system is fixed and equal n.  According to the postulate of equal probability, all microstates of the system are equiprobable.  The probability of finding the particle in the volume V1 is equal to p  N1 / N  V1 / V , as volumes V1 and V are fixed, p is constant.  The probability of finding the particle in the other part of the volume V  V1 is q  1  p . can be determined using the Hence, the sought probability binomial distribution (1.1). In applying formula (1.1) to an ideal gas, the volume V1 is not important, and it was taken only in order to visualize the probability p for an individual particle to be in this volume. For very small k (k  0) and very big k (k  n) the probability w(k / n ; p) is very small: w(k / n ; p)  q n  0 ,

as q and p are less than one, and n is very big. At a certain intermediate value k of the probability w(k / n ; p) reaches maximum, to find which it is necessary to solve the equation: dwk n ; p   0. dk

(1.6)

In [1.6.1] it was shown that nmax  n0 , where n0 is the number of particles per unit volume, if they were distributed uniformly throughout the volume. This result means that the most probable 24

distribution is the uniform distribution of particles throughout the volume V1 . As the location of the volume V1 is arbitrary and can be chosen in any part of the volume V, we can conclude that the most probable distribution of particles in the volume is uniform distribution. Such a state of a closed system, by definition, is stationary and equilibrium. Therefore, this result can be formulated as follows: the most probable state of the system is its equilibrium state. Fluctuations The number of particles in a certain volume is not constant and changes in a small range with time. As the state of the system is described by the binomial distribution, the average number of particles in the volume V1 is equal to the mathematical expectation of the binomial distribution (see (1.3)):

 k  mk  p n .

(1.7)

It is said that the quantity fluctuates if its value fluctuates around the mean value. A measure of fluctuations is the standard deviation from the mean value, which coincides with the root of the variance and in case of the binomial distribution is (see (1.5).):

k 2



 n p q.

(1.8)

This equation shows that the standard deviation increases more slowly than the total number of particles in the system, whereas the mean value (1.7) increases proportionly to the number of particles in the system. Let us find the relative standard deviation:

k 2 k



25

q p

1 . n

From this expression it follows that the relative standard deviation decreases with an increase in the number of particles in the system. Taking into account that q  1  p , and p  V1 / V , we obtain:

k 2 k



1 V . 1 V1 n

(1.9)

For V1  V the relative number of fluctuations approaches zero and for V1  V it is equal to zero, as in the whole volume of the total number of particles is fixed and there are no fluctuations in the number of particles. As V1 decreases the relative number of fluctuations increases according to (1.9). For V1  V we can neglect one as compared to V /V1 , and (1.9) is written as:

k 2 k



V V1

1 . n

From this formula it follows that the relative role of fluctuations increases as the area, in which these fluctuations are considered, decreases. For example, if we take the area where on average there are only a few particles, the relative value of fluctuations is a very significant fraction of the number of particles. If the area is so small that on average there are only 10 particles, the relative standard deviation reaches about 1/3. If under normal atmospheric conditions we take a volume of 1mm3, it contains on average 2.71016 particles, and the relative standard deviation is less than 10-8, which is a very small value. Therefore, in macroscopic systems with a very large number of particles, statistical fluctuations are insignificant, and with high accuracy it can be assumed that the values are equal to their mean values. 1.2.6. Gauss-Laplace distribution. When n   the binomial probability distribution transforms into the Laplace-Gauss formula. 26

Applying the Stirling formula for factorials, the expression (1.2) can be reduced to the following form (see APPENDIX 3): k  w ; p   n 

 1 e 2npq

 k n p 2 2n p q

.

(1.10)

This is the Laplace-Gauss formula. Taking into account the fact that this distribution coincides with the binomial distribution, it can be stated that the mathematical expectation and variance of the Laplace-Gauss distribution are calculated by the formulas (1.3) and (1.5). Taking into account these expressions, the formula (1.10) can be rewritten as follows:  1 k  w , p   e  n   k 2

k  mk 2 2 k2

.

(1.11)

Thus, for large n (namely, n>10) instead of (1.1), we can use expressions (1.10) or (1.11). Note that here the random variable k, as well as in the binomial law, is discrete. 1.2.7. Gaussian distribution. The fact that the value of a physical quantity measured in the experiment is a random variable, enables us to use the probability theory for processing of the experimental data. Indeed, the results of several measurements of electrical pulses are described by the Gaussian distribution: wx  

1

 x 2



e

 x mx 2 2 x2

,

(1.12)

where m x is the mathematical expectation of the variable x ,  x2 is the varianceof the variable x . Function w(x) determines the probability that the value of the measured variable x is in the interval x, x  dx (and this probability is w( x) dx ). 27

On the other hand, the system «measuring device – studied object» is affected by n uncontrolled (random) reasons, each of which is mutually independent deviations   and   , and p( )  p( )  1 . The combination of these deviations gives the error in the value of the physical quantity x  x  x  . This model up to its terminology coincides with the binomial model of the experiment, and the Gauss-Laplace distribution (1.11) is practically coincides with the Gauss law (1.12). The only difference between these formulas is that the distribution (1.11) contains a discrete random variable k, and the argument in the distribution (1.12) is a continuous random variable x. It should be kept in mind that the main characteristics of the continuous random variable are the mathematical expectation and variance, which are defined as follows:

mx   xw( x)dx ,

 x2   ( x  mx ) 2 w ( x)dx.

Using the Gaussian distribution (1.12), we can obtain a formula for the distribution of molecules of an ideal gas by their velocity and kinetic energy – the so-called Maxwell distribution (see APPENDIX 4). 1.3. Experimental technique To verify the binomial distribution experimentally we use the experiment with coin tossing. In a closed vessel, there are 12 coins. To determine the probability of k «eagles» of the total number n of coins, we must throw n coins simultaneously many times, noting in each experiment, the number of coins dropped out «eagle». Let the experience be made N times. After determining the number of experiments N (k ) in which k coins dropped out «eagle», we can find the relative frequency N (k ) / N of falling of k «eagles» for simultaneous tossing of n coins. In this experiment k may take the following values: k = 0, 1, 2, 3, …, n. Obviously, the more experiments will be carried out, the better result of the experiment we will get, and the probability determined experimentally for N   will coincides with the theoretical probability from (1.1): 28

N k  . k  w ; p   lim N   n N  

As the number of experiments is always limited, we suppose that

 N k  . k w ; p   N n 

1.4. Procedure of work fulfillment 1.4.1. Make from 100 to 150 experiments, counting in each experiment the number of coins dropped «eagles» up (or «tails» – by the choice of the experimenter). Write down the experimental results in Table 1.1. If the result of each test is marked by its number over the interval corresponding to the value k, by the end of the experiment we will get the histogram. The number of coins dropped «eagle» up № 1 2 … 33

№ 34 35 … 66

k

k

№ 37 38 … 100

Table 1.1

k

1.4.2. Determine the frequency N (k ) of occurrence of the random variable k and the experimental probability wexp  N (k ) / N from the histogram. Write down the results in Table 1.2. 1.4.3. Calculate theoretical probability distributions by formulas (1.2), (1.10), (1.11), using the following tables: Determination of  k  and the mean square error S k by the experimental data k 0 1 … 12

N(k)

wexp=N(k)/N

k wexp

∑=

∑=

=

29

k=k-

Table 1.2

k2 wexp

Sk2=

Table 1.3

Calculation of the binomial distribution (BD) for n=12 k

k C12 

12 k

k

12! k! (12  k )!

k 1 1 wБР  C12      2  2



k C12 4096

0 1 … 12 Тaблицa 1.4 Calculation of the Laplace-Gauss distribution (LG) for n=12 ( mk  6 ,  k2  3 ) z 

k

k

 6 6

2

e z

wЛГ 

1 e  z  0,23e  z 3  2

0 1 … 12

To calculate Gaussian distribution use  k  and S k values determined from the experimental data from Table 1.2. Table 1.5 Calculation of the Gaussian distribution (G) for  k  =…, S k =… х

z

x   k  2 2S

2 x

e z

wГ 

Sx

1 e z 2

0 1 … 12

1.4.4. Using the data from 1.2 – 1.5 make a summary table for comparison of all distributions and experimental data. Comparison of probability distributions k= x 1 0

Wexp 2

WBD 3

30

WLG 4

Table 1.6

wG 5

1 1 2 … 12

2

3

4

5

1.4.5. Using the data from Table 1.6 plot wBR functions wBD, wLG, wG on the same figure. Put experimental data wexp on the same figure. 1.4.6. Analyze the results and draw a conclusion about the reasons for these differences. Questions for self-assessment 1. What do the notions «a discrete quantity» and «a continuous quantity» mean? 2. What is the difference between the binomial distribution, the Laplace-Gauss distribution and the Gaussian distribution? 3. Where is the binomial distribution used in physics? 4. Give the definition of the mathematical expectation and variance of the random variable. 5. What are the mathematical expectation and variance of the random variable obeying a binomial distribution? 6. Why is it necessary to shake the cup with coins before each tossing? 7. What is the reason for the discrepancy of the experimental results from the theoretical probability distribution? References 1. Matveev A.N. Molecular Physics: a textbook for physical specialties of universities. Ed. 2nd, Revised and complementary. – M.: Higher School, 1987. – 360 p. 2. Hudson D. Statistics for physicists. – M.: Mir, 1979. – 295 p. 3. Bocharov P.P., Pechinkin A.V. Probability theory. Mathematical statistics. – M.: Gardarika, 1998. – 326 p.

31

Laboratory work →2 CALIBRATION OF THERMOELECTRIC THERMOMETER 2.1. The purpose of work: study of the operating principle of thermoelectric thermometer and its graduation.

2.2. A brief theoretical introduction The contact between two wires made from different materials causes appearance of the electron exchange between them, which leads to the appearance of the contact potential difference. Its value depends on the nature of the contacting conductors and contact temperature. Appearance of the contact potential difference is due to two reasons: 1) the difference in the electron work function of various metals; 2) the difference in the electron gas density in different metals, i.e., the number of electrons per unit volume of the metal. Consider the circuit (Fig.2.1) consisting of two welded together conductors 1 and 2 made from different metals.

Fig. 2.1

Fig. 2.2

32

While the temperature of junctions is the same, the contact potential differences arising in junctions A and B are equal. The equivalent scheme of the circuit is shown in Fig. 2.2, where the contact emf of junctions is shown as two elements with equal emf switched in the opposite directions: E A  EÁ .

(2.1)

The current in this circuit is equal to zero. When heating one of the junctions, the potential differences arising in the contacts will not be equal, so in this circuit resultant emf is different from zero, which gives rise to a current. This emf is called the thermoelectric force (thermal emf). The phenomenon was discovered in 1821 by T.I.Zeebekom and bears his name. It is widely used for measuring temperatures in devices called thermoelectric thermometers. Thermocouple Thermometer is a circuit containing two junctions of dissimilar metals in one of the conductors of which a millivoltmeter is included. Fig. 2.3 is a diagram of the thermoelectric thermometer. It consists of two conductors 1 and 2 from different metals the ends of which are welded to form junctions A and B.

Fig. 2.3

When measuring temperature, one junction is kept at a constant temperature (for example, is placed into the Dewar vessel with melting ice, t = 0 °C), the second junction is in thermal contact with 33

the tested body. These thermometers have the advantage that they can measure both very high and low temperatures, which cannot be done with the help of conventional liquid thermometers; moreover, they are more sensitive and less inertial. The value of thermal emf of the thermoelectric thermometer is determined by the difference in thermal emf between junctions A and B. If A junction is at t = 0 °C, the dependence of the thermal emf on the temperature t of «hot» junction can be represented as: E  at  bt 2  ct 3  ... ,

(2.2)

where a b and с are constants dependent on the kind of materials from which the thermoelectric thermometer is composed, t is the temperature of the «hot» junction in °С. To use the circuit containing two junction of dissimilar conductors as the thermometer, it is necessary to calibrate it. For calibration some preliminary known values of temperature, for example, the temperature of melting ice, boiling water, melting of pure metals are usually used. During calibration one junction is thermostated (i.e., kept at a constant temperature) in the Dewar with melting ice and the other is immersed into baths with known temperatures. Using a thermoelectric thermometer for precise temperature measurements, it is better to measure the electromotive force in the circuit rather than the current in it. This is due to the fact that the thermal emf depends on the type of metals forming junctions and their temperature, whereas the current flowing in the circuit also depends on the resistance of the measuring instrument, the resistance of connecting wires and the internal resistance of junctions. To characterize thermoelectric properties of a pair of conductors we introduce the concept of differential thermal emf , which is also called thermocouple constant and is equal to thermal emf arising when the temperature difference between the junctions in 1°C:

  dE / dt ,

(2.3)

 depends on the type of this pair of conductors and the temperature. APPENDIX 2 has the table of thermal emf values table for some of the most commonly used metal pairs. 34

Within small changes in temperature of the «hot» junction (for thermal emf values table for some of the most commonly used metal pairs. Within the small changes in temperature of the «hot» junction (for copper-constantan it can roughly be considered in the range of 100 °C) the temperature dependence of the thermal emf is linear: (2.4)

E1  at .

In this case, the thermocouple constant α depends only on the choice of metal vapors. For more accurate calibration, the dependence of the thermal emf on the temperature can be written as a quadratic function: (2.5)

E2  at  bt 2 .

The task is to remove the experimental dependence of thermal emf on the temperature of «hot» junction and evaluate which of the dependencies – (2.4) or (2.5) – is in better agreement with the experimental data 2.3. Experimental setup and measurement technique 2.3.1. The experimental setup is shown in Fig. 2.4. 1 is Dewar, 2 and 3 are «cold» and «hot» junctions of the thermoelectric thermometer, respectively, 4 is a vessel with a heated liquid, 5 is the melting ice. Constanta n Thermome ter

Voltmeter B7-21

Glycerol

Dewar container (ice+water)

Fig. 2.4. Experimental setup

35

The free ends of the thermocouple are connected to the terminals of the voltmeter. 2.3.2. Glycerin is used as the heated liquid. The boiling point of pure glycerol is 240 °C. Therefore, it is possible to change the temperature of the «hot» junction of a thermoelectric thermometer in the range from room temperature to 240 °C. 2.4. Experimental procedure 2.4.1. Prepare a voltmeter to work. 2.4.2. Determine the scale division the thermometer. 2.4.3. Fill the Dewar vessel with melting ice and place one of the junctions of the thermoelectric thermometer into it. 2.4.4. Measure thermal emf for the case when the liquid, into which the so-called «hot» junction is placed, is at room temperature. 2.4.5. Turn on the heater and measure thermal emf with an interval of 10 °C in temperature changes in the «hot» junction. Write the results down in Table 2.1. Results of measurements of thermal emf for different temperatures of «hot» junction

t hot and calculations of a coefficient as a

function of E1  at (2.4), № 1 2 … n .

ti hot , 0C

Table 2.1

Ei exp , мВ

d i  Ei exp  Ei calч xi2

xi y i

n

x y i 1

i

i



n

x i 1

2 i

Ei cal

di



2.4.6. Using the least squares method, calculate the coefficients in (2.4) based on the experimental data (see. APPENDIX 1, formula A3). 2.4.7. It is convenient to calculate parameters in (2.5) E2  at  bt 2 using Table 2.2. The dependence (2.5) can be written as: 36

E  a  bt , t

i.e. it can be reduced to a linear dependence y  A  Bx , where y

E, A a, B  b, x t . t

Then, to calculate coefficients A and B we can use formulas (A 2), given in the APPENDIX 1. Using coefficients A and B calculate theoretical values of yi ðàñ÷  A  Bxi . Find the values of Ei ðàñ÷  yi ðàñ÷  t i and write them down in Table 2.2. Data for calculating parameters of the function E2  at  bt 2 (2.5), d i  Ei exp  Ei cal №

Ei exp / t i

ti xi

xi2

xi y i

Ei cal

Table 2.2

di

yi

1 2 . n n

x i 1

i



n

y i 1

i



n

x i 1

2 i



n

x y i 1

i

i



2.4.7. Put the experimental points on the graph of the dependence of thermal emf on temperature, draw the curves (2.4) and (2.5), using the obtained values of Ei cal . 2.4.8. Estimate which of the two dependencies is in better agreement with the experimental data. Questions for self-assessment 1. Explain the principle of operation of thermoelectric thermometer. Which physical phenomenon is the basis of its work? 2. What factors lead to the appearance of the contact potential difference? 3. What are the advantages and disadvantages of the method of temperature measurement with a thermoelectric thermometer? 4. What determines the range of temperature measurement for liquid thermometers and thermoelectric thermometer?

37

References 1. Practicum in physics. Mechanics and molecular physics: textbook / Belyankin A.G., Motulevich G.P. et al.; Ed. Iverenova V.I. – M.: Nauka, 1967. – P. 352. 2. Laboratory works in physics: Textbook / Goldin L.L., Igoshin F.F., Kozel S.M., et al.; Ed. Goldin L.L. – M.: Nauka. Main publishing house of Physical and mathematical literature, 1983. – P. 704. 3. Kay J., Lebi T. Tables of Physical and Chemical Constants / J. Kay, T. Lebi. – M., 1962. 4. Kalashnikov. S.G. Electricity / Kalashnikov. – S.G.: Nauka, 1970. – 666 p. 5. Matveev A.N. Electricity and Magnetism / A.N. Matveev. – M.: Higher School, 1983. – 463 p. 6. Zeidel A.N. Errors of measurements of physical quantities / A.N. Zeidel. – SPb.: Lan, 2005. – P. 106.

38

Laboratory work →3 DETERMINATION OF BOLTZMANN CONSTANT 3.1. The purpose of work: 3.1.1. Study the experimental method of determining the Boltzmann constant. 3.1.2. Determine the value of the Boltzmann constant and estimate the measurement error.

3.2. A brief theoretical introduction The ideal gas model. This is the simplest model of the gaseous state of matter. In this model, we neglect the potential energy of interaction between the molecules of gas in comparison with the kinetic energy of chaotic motion. This approximation is possible if the average distance between the molecules is much greater than their diameters. However, collisions of molecules with each other cannot be ignored. It is collisions between molecules that lead to the randomness of their movement and thermodynamic equilibrium. Such an approximation also corresponds to the mechanical model of elastically colliding balls with small size (compared to the size of the vessel). The basic equation of kinetic theory of gases establishes a relationship between the molecular quantities, i.e., quantities relating to a single molecule, and the amount of pressure that characterizes the gas as a whole – the macroscopic quantity directly measured in the experiment: 2 m 2  2 (3.1) p n o  n   const  , 3

2

3

where p is the gas pressure on the walls of the vessel, n is the number of molecules per unit volume, m0 is the mass of the molecule,   2  is the mean square of the velocity of thermal motion of molecules, 39

  пост  is the average kinetic energy of the translational mo-

tion of the molecule. The number of molecules per unit volume is equal to the ratio of the total number of gas molecules N in the vessel to its volume V : n  N V , then (3.1) can be rewritten as: pV 

2 N   const  . 3

(3.2)

Therefore, the product of the gas volume to its pressure is numerically equal to 2/3 of the translational kinetic energy of chaotic motion of gas molecules in this volume. The total number of molecules is: N  mN A / M ,

where m / M is the number of gas moles, N A is the Avogadro's number, i.e. the number of molecules in one mole of any substance. Substituting this expression in (3.2), we obtain: pV 

2 m N A   const  . 3M

(3.3)

Compare this equation with the empirical equation of state of Mendeleev – Clapeyron: m (3.4) RT . pV  M

We obtain:

  const  

3 RT . 2 NA

(3.5)

Thus, the average kinetic energy of the random motion of a single molecule of gas is directly proportional to its absolute temperature. The quantities R and N A are universal constants. Their ratio: 40

R k NA

(3.6)

is also a universal constant and is called the Boltzmann constant. If R  8,31 J (mol  К ) , N A  6,02 10 23 mol 1 , then k  1,38  10 23 J / К . Using the Boltzmann constant, we obtain for the average kinetic energy of a molecule: 3 (3.7)   const   kT . 2 Therefore, the average kinetic energy of random translational motion of molecules of an ideal gas is directly proportional to its absolute temperature and is a measure of intensity of thermal motion of molecules at a given temperature. The formula (3.7) reveals the molecular-kinetic meaning of the concept of temperature: The temperature of a body is a quantitative measure of the energy of thermal motion of molecules that make up the body. Substituting (3.7) into (3.1), we can transform the basic equation of the kinetic theory of gases to the form: (3.8)

p  n kT .

This ratio connects the gas pressure with the number of molecules n per unit volume, and absolute temperature T. The coefficient of proportionality between them is the Boltzmann constant. Equation (3.8) is valid not only for pure gases but also for gas mixtures. In the case of non-interacting mixture of ideal gases, pressure p is the total pressure, and n is the number of molecules of the gas mixture per unit volume. Indeed, as n   ni , then p   ni kT   pi . i

i

41

i

(3.9)

The total pressure of the gas mixture is the sum of the partial pressures of its components (Dalton's law). Partial pressure of the i-th component of the gas mixture is the pressure of the component at the same values of temperature and volume, as those of the mixture. Equations (3.8) and (3.9) can be used for pure vapors and mixtures of chemically inert vapors, provided that their partial pressures are lower than the corresponding pressures of saturated vapors. For saturated vapors ratios (3.8) and (3.9) are not satisfied. 3.3. Description of the experimental setup and measurement technique 3.3.1. The experimental setup for measuring the Boltzmann constant is shown in Figure 3. The setup consists of the vessel 1 of volume V 1 with the cover 2. Before starting the experiment it is necessary to ventilate the container. For this purpose we must rotate the valve 4 so that the vessel is connected with the pump 3. The vessel is hermetically sealed with plug 8. Through the vent in the vessel it is connected with the inclined micromanometer. The plug has a needle 5, through which the 1 cm3 of easily evaporating liquid (e.g. acetone) is injected by a syringe. Molar mass M, density  and vapor pressure pí at 20 °C for some liquids are given in the APPENDIX to this laboratory work.

1 – vessel, 2 – cover, 3 – pump, 4 – valve, 5 – needle, 6 – base, 7 – hose, 8 – plug Fig. 3. Setup for determining the Boltzmann constant

42

In the evaporation of liquid the vapors create excessive partial pressure p measured by the inclined micromanometer. The description of the inclined micromanometer is given in APPENDIX 1 at the end of the book. The experimental setup is fixed on the base 6. The resulting partial pressure of liquid vapors is equal to the excess pressure p in the vessel: p  nkT ,

then k  p nT .

Taking into account that n  N / V is the number of vapor molecules of the liquid divided by the volume of the vessel, we get: N

Vliq m    N AVliq , m0 M / N A M

(3.10)

where  is the density of the injected liquid, for acetone

  0,7893 г / см 3 , m – is mass of the injected liquid, Væ is the volume of liquid injected into the vessel with a syringe, m0 is the mass of one molecule, M is the molar mass of liquid, M  58,08  10 3 kg / mol . Then to determine the Boltzmann constant we obtain: k

MV  p ,

 N AVliqТ

(3.11)

where p is the excess pressure in the vessel, determined by the inclined micromanometer,  p  kì  g  l (Pa),

k ì – is the inclination coefficient of the micromanometer, l is the reading of the manometer. Make the experiments 3 times. Write down the results in Table 3.1. 43

Results of measurement of excess pressure of liquid vapor №

Liquid

Vliq , m3



Table 3.1

p, Pa

l

1 2 3 Table 3.2

Calculation of the Boltzmann constant k №

k,

p, Pa

Jou  10 23 K

 k , Jou  10 23 K

k , Jou  10  23 K

k 2 , Jou 2  10  46 K2

1 2 3

3.3.2. Estimate the error of the result taking into account the random error and the instrument error according to the formula: 2 2 . k const  k ran  k inst

(3.12)

Calculate the random error k ran using the standard method. For one of the measurements estimate the instrument error k inst , using the method of finding errors for indirect measurements: 2

k inst

 Vж  V  k     V    Vж

2

2

2

 T   l        , T    l  

(3.13)

where V is the error in measuring the volume of the vessel (given on the device); Vliq is the instrument error of the syringe equal to half of the lowest division; T is half of the lowest division of the thermometer; К, l is the error of measurement on the scale of the pressure gauge equal to half of the division. 44

Estimate the relative error:

 k  

k full k

.

Present the final result as: k   k   k full ,

J . К

Questions for self-assessment 1. What is an ideal gas model? Under what conditions is a real gas close to the ideal? 2. Write the basic equation of the kinetic theory of gases, explain its meaning. 3. What is the physical meaning of temperature and Boltzmann constant? 4. Derive the equation of state of an ideal gas. 5. What is called a partial pressure of the component in the gas mixture? Formulate Dalton's law for gas mixtures. 6. Why it is necessary to ventilate the vessel before the experiment? Is it necessary to remove the vapors of liquid before starting a series of experiments with the other liquid? 7. Explain the construction and function of operation of the inclined micromanometer. 8. What are the units of measurement of pressure in the SI? What non-SI units of pressure do you know? 9. What is the relationship between pressure and temperature of the gas in the vessel? 10. Measurement of what quantity makes the largest contribution to the total error of the result? References 1. Matveev A.N. Molecular Physics: A Textbook for physical faculties of universities. – 2 nd ed., Revised and ext. / A.N. Matveev. – M.: Vysshaya shkola, 1987. – P. 360. 2. Kikoin A.K., Kikoin I.K. Molecular physics. – 2nd ed. / A.K. Kikoin, I.K. Kikoin. – M.: Nauka, 1976. – P. 480. 3. Savelyev I.V. The course of general physics. Bk. 3: Molecular Physics and Thermodynamics / I.V. Savelyev. – M.: Astrel. AST, 2003. – P. 208. 4. Laboratory works in physics: textbook / L.L. Goldin, F.F. Igoshin, S.M. Kozel, et al.; Ed. Goldin L.L. – M.: Nauka. Main publishing house of Physical and mathematical literature, 1983. – P. 704.. 5. Tables of physical quantities. Reference book. / I.K. Kikoin, I.S. Grigoriev, Yu. Danilov et al.; Ed. I.S. Kikoin. – M.: Atomizdat, 1976. – P. 1006. 6. Zeidel A.N. Errors of measurements of physical quantities /

A.N. Zeidel – SPb.: Lan, 2005. – P. 106. 45

Laboratory work →4 DETERMINATION OF SPECIFIC HEAT OF TIN CRYSTALLIZATION (MELTING) AND CHANGES IN ENTROPY DURING CRYSTALLIZATION 4.1. The purpose of work: 4.1.1. Determination of specific heat of tin crystallization (melting). 4.1.2. Determination of the change in tin entropy during crystallization.

4.2. A brief theoretical introduction 4.2.1. Phase transitions. Melting and crystallization. Part of a thermodynamic system, homogeneous in its macroscopic properties, is called a phase. Phases are separated from each other by physical interfaces. An example of two phases of the same substance is water with floating pieces of ice in it. Under certain conditions different phases of the same substance may be in equilibrium with each other, touching each other. Equilibrium occurs only in a certain temperature range, and each temperature T corresponds to a certain pressure p. Thus, equilibrium states of two phases are represented in the diagram  pT  by the line p  f T  . This diagram is called a state diagram (see Figure 4.1). Transition from one phase to another is usually accompanied by the absorption or emission of a certain amount of heat, which is called the latent heat of the phase transition or heat of transition. Such transitions are called phase transitions of the first type. Phase transitions of the first type are melting and crystallization, evaporation and condensation, sublimation and condensation. Melting, boiling and sublimation curves divide the coordinate plane (pT) into three areas (see Figure 4.1). Any point in one of these areas shows the corresponding uniform state of matter. Curves 1, 2, 3 are lines of phase equilibrium. The point T in which all three phases of the substance are in dynamic equilibrium with each other is called a triple point. 46

Point K is a critical point.

1 – sublimation and condensation, 2 – evaporation and condensation, 3 – melting and crystallization Fig. 4.1. Phase diagram

An example of the phase transition of the first type is the process of melting and the inverse crystallization process. The substance can transfer to the solid state both from the liquid and gaseous state. In both cases, this transition will be accompanied by a change in the symmetry of states. Consider the transformation «liquid – solid body». The process of formation of a solid body during liquid cooling proceeds at a certain temperature Tk – the temperature of crystallization. This is explained by the fact that the potential energy of crystallization or interaction between molecules and atoms decreases and is released as heat, which is the heat released by the «liquid – crystal» system to the environment. This heat is called the heat of crystallization. The heat released as a result of crystallization of a unit mass of the substance, is called the specific heat of crystallization. The reverse transformation – melting – also occurs at a constant temperature Tmel and is accompanied by energy absorption. The absorbed energy in this case goes on the destruction of the crystal lattice. For a given pressure the crystallization temperature Tk is equal to the temperature of the melting point Tmel . The amount of energy required for the unit of mass of the substance to be transferred from solid to liquid state at the melting tem47

perature, is called the specific heat of melting. The specific heat of melting is numerically equal to the specific heat of crystallization λ. Let us heat in an electric furnace the tested substance (tin) together with the crucible, where it is placed, to a temperature above the melting temperature. Then, let us remove the crucible from the furnace. The tin will start to cool in the air due to natural convection. After some time such conditions will be formed that the temperature in all points of the tin will decrease at the same rate. This condition is fulfilled the better, the greater is the thermal conductivity of the object to be cooled and the less is intensity of heat transfer to the environment at a constant temperature Tenv . This cooling process is called regular thermal regime of the first type. The amount of heat  Q , lost by a heated body in a time d during cooling is found using Newton's law for heat transfer:  Q   Tsur  Tenv   S  d .

Here  is the heat transfer coefficient, which depends on the shape of the heated body, the quality of its surface and physical parameters of the environment (air density, viscosity, etc.). The heat transfer coefficient  is numerically equal to the amount of heat lost by a unit of surface of the body per unit of time at Tsur  Tenv  1K ,    Vt / m 2 К ; Tsur is the surface temperature of the heated body; Tenv is the temperature of the environment; S is the area of the heated body. The equation of thermal balance for tin cooling is written taking into account the fact that the heat released during cooling of tin with the crucible by the temperature dT is removed to the environment during time d :









 cO1mO  cT mT dT  q1 Tsur  Tср d . Here cO1 – is the specific heat of liquid tin, J / (kgK); mo is tin mass kg; 48

(4.1)

cT and mT are specific heat capacity and the mass of the crucible, respectively, J/(kgK) and kg; dT – body temperature change over time d , K; q1 is the rate of heat transfer from the liquid tin to the environment, J/(sK); Tenv is the ambient temperature (air temperature in the laboratory), K;

Tsur is the surface temperature of the cooled body, K. The rate of heat transfer q1 , i.e. the amount of heat lost by the

heated body per a unit of time at Tsur  Tenv  1K depends on the surface area of the cooled body and the heat transfer coefficient α. As the first approximation, q1 can be considered constant for a certain form of the crucible and a small range of variation in the body temperature. Since the thermal conductivity of the tin is rather high and its temperature in the point of thermocouple location is not very different from the temperature of the outer surface, we can assume that T  Tsur . Taking into account this equation, we integrate equation (4.1) by the method of separation of variables. To do this, let us write it in the form: dT q1 (4.2) d .  T  Tenv

cO 1mO  cT mT

Let us use the notation: k1 

q1 . cO1mO  cT mT

(4.3)

k1 has the following unit of measurement in SI

k1   q / cm  1 c 1 ; k1 characterizes the rate of temperature change of the cooling

system and is called the rate of cooling. 49

To solve the differential equation (4.2) it is necessary to set the initial condition T  T0 for   0 . Integrating the right and left side of the equation (4.2), we obtain: (4.4) ln T  Tenv   k1  ln C , where C is the constant of integration. From the initial condition we find this constant: ln T0  Tenv   ln C and C  T0  Tenv . Then the expression (4.4) can be written as follows: or

ln T  Tenv   ln T0  Tenv   k1 ,

T  Tenv  T0  Tenv  e

 k1

.

(4.5) (4.6)

As it can be seen from (4.6), when the heated body is cooled, its temperature decreases exponentially with time. Such cooling rate is preserved as long as the temperature of tin reaches the crystallization temperature Tk . This temperature will not change till full crystallization of tin, although heat continues to release to the environment. When the crystallization process is over, the temperature of the crucible with tin starts to decrease again, following the exponential law: (4.7) T  Tenv  Tk  Tenv   e  k2 (  k ) , where q2 , (4.8) k2  cO 2 mO  cT mT

Tk is the tin crystallization temperature; k 2 is the rate of cooling of a solid tin with the crucible; cO 2 is the specific heat of solid tin. The curve of temperature dependence of time for tin cooling has the form shown in Fig. 4.2. The section AВ corresponds to cooling of liquid tin, BC is tin crystallization section, SD is cooling of solid tin. 50

Т, К

Fig. 4.2. Curve of tin cooling

, с

A small temperature minimum is observed near point B, and near point C there is a small area where the cooling rate is less than k 2 . Violation of the regular cooling regime near point B is caused by the occurrence of a metastable state (supercooled liquid state when the temperature falls below the crystallization temperature Tk and crystallization may not occur), which depends on the purity of the substance. The amount of heat released during crystallization of tin is: Q   mO ,

(4.9)

where  , the specific heat of crystallization, is numerically equal to the specific heat of melting. 4.2.2. Derivation of the formula for determining the specific heat of crystallization. The heat emitted during crystallization is equal to the amount of heat released by the body during crystallization to the environment:





Q  q Tk  Tср   k   н  ,

where q  ronment;

(4.10)

q1  q 2 is the average rate of heat release to the envi2

51

 í and  k are the moments of time corresponding to the beginning and end of tin crystallization. Equating the right sides of (4.9) and (4.10), we obtain:

mO  q Tk  Tср   k   н  . Hence: 





q Tk  Т cp   k   н 

.

(4.11)

mO

The expression for q can be found using formulas (4.3) and (4.8): (4.12) q1  k1 cO1mO  cT mT , q2  k 2 cO 2 mO  cT mT .

Hence, q

(4.13)

1 k1 cO1mO  cT mT   k 2 cO 2 mO  cT mT . 2

The final expression for  is written as 

Tk  Tñð  k   í  2 m0

k1 c01m0  cT mT   k 2 c02 m0  cT mT .(4.14)

k1 , k 2 ,  í ,  k are found from the graph, constructed based on

the experimental data and shown in Fig. 4.3. To do this, we write the equation (4.5) for the moment when the crystallization starts  í :









(4.15)

ln Tk  Tср  ln T0  Tср  k н .

From (4.15) we get: k1 







ln T0  Tср  ln Tk  Tср

н 52

.

(4.16)

Similarly, writing (4.5) for the end of crystallization  k and a moment of time  2 during cooling of solid tin, we have:

   ln T2  Tср   ln T

  Tср   k  2 .

ln Tk  Tср  ln T0  Tср  k 2 k , 0

2

Subtracting the second equation from the first one, we obtain the following formula for k 2 : k2 



 

ln Tk  Tср  ln T2  Tср

 2  k

.

(4.17)

In this laboratory work as a temperature gauge we a thermoelectric thermometer (see lab. No. 2. The principle of operation of thermoelectric thermometer). Thermo-emf E is proportional to the temperature difference between ‘hot’ T and ‘cold’ Tñð junctions, so the cooling curve for tin can be drawn in variables E and  . If the same curve is plotted in semi-logarithmic coordinates, ln E and  (Fig. 4.3), the sections AB and CD will be strasight lines, which will enable us to more accurately determine the beginning  í and end  k of crystallization. Thus, to determine the rates of cooling k1 and k 2 we must draw the dependence of ln E on  Fig. 4.3.

Moments of time  í and  k correspond to points of intersection of lines AB and CD with the crystallization line BC.

0

н

к

2

, c

Fig. 4.3. Tin cooling curve in semi-logarithmic variables

53

Since thermal emf in the first approximation is proportional to the temperature difference at the ends of junctions of the thermocouple, formulas (4.16) and (4.17) for k1 and k 2 calculation are written as:

k1  k2 

ln E0  ln E k

,

(4.16a)

ln E k  ln E 2 .  2  k

(4.17a)

í

4.2.3. Derivation of the formula for entropy change during crystallization. Entropy characterizes probability of the macrostate of a physical system and is determined as

S  k  ln W ,

(4.18)

which is called the Boltzmann formula. Here k is the Boltzmann constant; W is the statistical weight (thermodynamic probability), i.e. the number of microstates by which this macrostate is realized. Changes in the entropy in thermal interactions are determined based on the measurement of the reduced amount of heat. According to the thermodynamic definition of entropy, it’s increment is equal to the reduced amount of heat, i.e. to the ratio of the elementary amount of heat received by the system to the absolute temperature T at which this process occurs: dS  Q T .

(4.19)

The total change in entropy in a reversible transition of a system from state A to state B is determined by integrating the expression (4.19):  Q . (4.20) SB  S A   T  54

During crystallization tin releases heat to the environment, i.e. Q  0 , so tin entropy decreases. The change in entropy of tin during crystallization is determined by the formula:   mO (4.21) S  S B  S A  0. Tk Since entropy, to some extent, can be considered as a measure of disorder in the system, the decrease in entropy of the tin during crystallization corresponds to the transition of the system into a more ordered state. 4.3. Description of the setup 4.3.1. Materials and devices: a crucible with tin, a furnace, a measuring thermocouple, a stopwatch, a digital voltmeter. 4.3.2. The setup (Fig. 4.4) consists of a crucible with tin fixed at the end of the rod The crucible can be put into an electric furnace and removed from the furnace when the tin is melted.

1 – a crucible with tin, 2 – a tripod, 3 – a heater, 4 – a vessel with glycerol, 5 – a measuring thermocouple Fig. 4.4. A setup for determining the specific heat of tin melting

One junction of the thermocouple is in glycerol at room temperature, and its «hot» junction is placed in the molten tin in the crucible. 55

The thermocouple is connected to a universal digital voltmeter to measure the thermo-electromotive force (thermal emf) in the thermocouple circuit. Readings of the voltmeter are proportional to the temperature difference between the junctions of the thermocouple: E   T  Tenv  ,

4.22)

where α is the thermocouple constant. 4.4. Procedure of work fulfillment 4.4.1. Write down the data necessary for calculations in your notebook: cO 1 = 267.9 J/(kgK) is the specific heat of liquid tin, corresponding to a temperature of 240 °C; cO 2 = 246.6 J/(kgK)) is the specific heat of solid tin, corresponding to a temperature of 230 °C; cT , mT and mO are the specific heat of a porcelain crucible, the mass of the crucible and tin (given on the label near the Setup); Tk = 231.9 °С is tin crystallization temperature;

Tñð is the air temperature as indicated by the reference thermometer hanging on the wall in the laboratory. 4.4.2. Place the crucible with tin in the holder above the furnace so that the crucible could enter the central opening of the muffle furnace. 4.4.3. Switch on the setup. 4.4.4. While tin is heated, prepare a table for recording the results of the experiment, making the following columns in it:  the number;  the time of seconds;  thermal emf in millivolt;  logarithm of thermal emf, ln E . The table must have 50-100 lines depending on the time intervals used for measurements. 4.4.5. When the voltmeter shows 20 mV, turn off the oven and take the crucible holder away from the heater, placing it over the table at a height of 15-20 cm. 56

4.4.6. Turn on the timer and start recording voltmeter readings every 10-20 seconds. Make measurements until the voltmeter has the value two times smaller than the value of thermal emf corresponding to the crystallization temperature. Write down the data in Table 4. Results of measurements of thermal emf E and values of ln E

, c

E, mV

Table 4

ln E

0 15 30 …

4.4.7. Draw the curve of tin cooling in variables E  f   . 4.4.8. Make the same graph in semi-logarithmic coordinates ln E  f   and determine time moments corresponding to the beginning  í and end  k of the crystallization process, and the value ln E k corresponding to the temperature of crystallization. 4.4.9. By the slope of sections AB and CD of the semi-log graph determine the rates of cooling k1 and k 2 using formulas (4.16a) and (4.17a). 4.4.10. Using table values of the temperature of tin melting and measured ambient temperature in the laboratory, find the specific heat of tin melting from formula (4.14). 4.4.11. Using formula (4.21) calculate the change in tin entropy during crystallization. 4.4.12. Compare the obtained value of  with its tabule value from the reference book. Questions for self-assessment 1. Tell about phase transitions of the first type. Draw a state diagram and explain it. 2. What is called the specific heat of melting? 3. In which phase is the substance during crystallization and how does its internal energy change? What part of it (kinetic or potential) changes? 4. How does entropy of the body change during crystallization? 5. How can we explain the existence of transitional areas on the cooling curve near points B and C in Fig. 4.1?

57

6. What errors occur in measurements? Explain the reasons for them. 7. What simplifying assumptions are made in the derivation of (4.14), and how will they affect the accuracy of the result? References 1. Matveev A.N. Molecular Physics: a Textbook for physical faculties of universities. – 2 nd ed., Revised and ext. / A.N. Matveev. – M.: Vysshaya shkola, 1987. – P. 360. 2. Kikoin A.K., Kikoin I.K. Molecular physics. – 2nd ed. / A.K. Kikoin, I.K. Kikoin. – M.: Nauka, 1976. – P. 480. 3. Savelyev I.V. The course of general physics. Bk. 3: Molecular Physics and Thermodynamics / I.V. Savelyev. – M.: Astrel. AST, 2003. – P. 208. 4. Laboratory works in physics: textbook / L.L. Goldin, F.F. Igoshin, S.M. Kozel et al..; Ed. Goldin L.L. – M.: Nauka. Main publishing house of Physical and Mathematical Literature, 1983. – P. 704. 5. Kortnev A.V., Rublev Yu.V., Kutsenko A.N. Practicum on Physics / A.V. Kortnev, Yu.V. Rublev, A.N. Kutsenko. – M.: Vysshaya shkola, 1963. – P. 516.

58

Laboratory work →5 DETERMINATION OF RELATION Cp/CV USING THE STANDING WAVE METHOD 5.1. The purpose of work: determination of the ratio of specific heats of gases by standing waves.

5.2. A brief theoretical introduction To characterize thermal properties of the gas, as well as any other body, the physical quantity called heat capacity is used. Heat capacity C т of a body is a quantity equal to the amount of heat needed to change its temperature by 1 K. If the body temperature does vary not by 1 K, but by dT degrees, this quantity is equal to: Cт  Q dT ,

(5.1)

where  Q is the amount of heat needed to change the body temperature by dТ degrees. C ò is measured in J/K. Heat capacity per unit mass of a substance is called specific heat capacity ñ , i.e. Cт Q . (5.2) с



m

m  dT

Specific heat capacity is measured in J /(kg·K). Heat capacity referred to one mole of a substance is called molar heat capacity Ñ : C







Q .

  dT

Here   M m is the number of moles of the substance, M is its molar mass. 59

(5.3)

The unit molar heat capacity in SI is J/(molK). It is obvious that

C  Mc .

(5.4)

Both specific and molar heat capacity characterize not the body but the material from which it is composed. The value of heat capacity depends on conditions under which the body is heated. Of greatest interest is heat capacity for the cases when the body is heated at constant volume or constant pressure. In the first case the heat capacity is called the heat capacity at constant volume CV , in the second – the specific heat at constant pressure C p . For heating of a given mass by 1K at constant pressure and at constant volume, different amounts of heat are required i.e.  Q   Q  , CV     Cp     dT V  dT  p

(5.5)

heat capacity C p of the ideal gas is greater than its heat capacity

CV . This is due to the fact that when V  const the heat energy is spent only on the change in internal energy, i.e. on the increase in temperature, whereas when p  const some part of heat goes on gas work during expansion. The Meyer formula C p  CV  R shows the physical meaning of the universal gas constant R. Universal gas constant R is numerically equal to the work done by one mole of an ideal gas at its isobaric heating by 1 K. The heat capacity of the substance is closely related to its internal structure and can be theoretically calculated under certain assumptions about its structure. According to the classical theory based on Boltzmann's law on the uniform distribution of energy over degrees of freedom, we have the following expression for the ratio of the heat capacities СP and СV: 

CР i  2 ,  CV i 60

(5.6)

where i is the number of degrees of freedom of gas molecules. As it follows from (5.4), the value of  is determined by the number of degrees of freedom of the molecule. The number of degrees of freedom of a mechanical system is the number of independent coordinates defining its position and configuration in space. For a monatomic gas i  3 , for a diatomic gas i  5 , for triatomic gas i  6 . However, formula (5.4) not always gives values coinciding with the experiment. For example, it cannot explain the heat capacity of chlorine. This this theory is not able to properly take into account the energy associated with internal movements in the molecule to which the law of uniform energy distribution over degrees of freedom is not always applicable. A very important deviation from the results of the theory is the fact that С Р СV depends on temperature, whereas, according to (5.4), it is for the gas with a given value of i is a constant. The discrepancy between the theory and the experiment shows that the idea of molecules as hard spheres, which move according to the laws of mechanics, is not correct. Now it is well known that molecules consist of interacting atoms and atoms have a complicated structure and consist of many even smaller particles, which also move in a complicated manner. The motion of atomic particles does not obey the laws of classical mechanics and is governed by the laws of quantum mechanics. Therefore, when we consider heat capacity of monoatomic gases, which is not affected by intra-atomic movement and corresponding energy, the classical theory of heat capacity is in good agreement with the experiment. However, in polyatomic molecules an important role is played by internal processes in molecules and atoms, to which, for example, vibrational degrees of freedom are related. In this case, the classical theory, not taking into account special quantum properties of atomic systems, only gives approximately correct results. The quantum theory gives a full explanation of all experimental data on the heat capacity. The ratio of heat capacities of gases   С Р СV plays an important role in the theory of ideal gases. As mentioned above, it is associated with a number of degrees of freedom of molecules. Moreover, 61

this quantity is included in the equation of adiabatic process (Poisson

equation): pV   const . This quantity is important, because knowing it we may avoid measuring CV , which is always difficult to do. This quantity can be obtained from the measured values C p and  , which is often done in practice. There are several methods used to determine С Р СV . The most convenient of them is the method based on measuring the speed of sound in the gas. From acoustics it is known that the speed of sound in the gas  is determined by the formula:

   RT M ,

(5.7)

where R  8,31J / mol  К  ) is the universal gas constant; T is gas temperature; M is its molar mass. If T and M are known, then finding the speed of sound  from equation (5.5), we can find  from the equation: 

M 2 . RT

(5.8)

To find the speed of sound we can use the equation:

  f  ,

(5.9)

where f is the frequency of sound vibrations in the studied gas; λ is the wavelength. To determine the wavelength  the method of standing waves is used. Standing waves arise as a result of vibrations observed in the superposition of two counter-propagating plane waves with the same amplitude. In practice, standing waves occur when waves are reflected from obstacles. The wave incident on the barrier and the reflected wave running towards it are superimposed and give a standing wave. If we take the 62

origin of coordinates in the point where the opposing waves have the same phase, and choose the time so that the initial phases were equal to zero, then the equation for the two plane waves can be written as follows:  x упад  a0 cos   t   ,    x yотр  a0 cos   t   ,  

(5.10)

where x is the distance from the source of vibrations to the observation point, y is the displacement of the vibrating point from the equilibrium,   2 /  is the cyclic oscillation frequency. The addition of these two waves gives:  x  x у  у пад  у отр  a 0 cos   t    a 0 cos   t         x  2a 0 cos    cos  t  A cos  t.  

(5.11)

Equation (5.11) shows that vibrations with frequency  arise in all points of the medium. These vibrations are called standing waves. The factor 2a0 cos x /   , independent of time, gives the amplitude of the resulting vibration. Thus, the amplitude of the vibration depends on the coordinate x , determining the position of the points in the medium. The points in which the amplitude is maximum are called antinodes. The points in which the amplitude is zero, do not take part in vibrations. Such points are called nodes. The amplitude A is maximal in the points where (5.12) cos x /   1 . The position of antinodes is determined by the condition: where k = 0, 1, 2...

 xk   k ,

63

(5.13)

Hence, coordinates of antinodes are: x k  k

f f    k  k  k .   2 f 2

(5.14)

In the tube of length l closed from both sides (phone and microphone) a pure resonance wave is formed when the length of the tube is equal to the integral number of half wavelengths, i.e. when l  k / 2 . In this case, the intensity of sound received by the microphone is maximal. The speed of sound is determined as follows:   f    f  2l k .

(5.15)

It is not convenient to find the value of k experimentally. Therefore, it is defined as follows: at a frequency f1 it is supposed to be k  k1 , at the other frequency f 2 the value of k will change and will be equal to k 2 , that is:   f1

2l and 2l   f2 k1 k2

  fi

2l . ki

Then the speed  can be determined as: 

2l  f 2  f1  . k 2  k1



2l  f i  f1  . k i  k1

(5.16)

It is easier to find the difference k 2  k1 than the absolute value. From (5.14) we get: f 2  f1  . (5.17)  k 2  k1 2l 5.3. The experimental setup and experimental technique 5.3.1. Description of the experimental setup. The block diagram of the setup used in this work is shown in Figure 5. Sound vibrations in the steel tube are excited by the telephone T and 64

are received by the microphone M. The membrane of the phone operates under the action of audio-frequency alternating current, sound generator GZ-34 is used as a source of variable emf. Frequency of the excited vibrations is measured by the frequency counter ChZ-32. The signal in the microphone is registered on the digital voltmeter V7-35. When a resonance arises (i.e. when l  k / 2, k  0, 1, 2, ... ), the digital voltmeter will show maximal value. Valves 1 and 2 are, respectively, used for gas filling and pumping. Pumping is made with a vacuum pump.

Fig. 5. Block diagram of the experimental setup

5.3.2. Procedure of work fulfillment. 5.3.2.1. Plug in the sound generator GZ-34, electronic frequency counter ChZ-32 and digital voltmeter V7-35. Let them get warmed for 5-7 minutes. 5.3.2.2. Under the supervision of the teacher or a laboratory assistant pump the gas, remaining from the previous experiment, out of the tube and fill it with air. 5.3.2.3. Slowly increasing the frequency of the signal supplied by the sound generator, obtain several, about ten, successive values of resonance frequency, marking the onset of the resonance by increased readings of the digital voltmeter V7-35. Repeat the experiment three times. Find the average value of the frequency corresponding to each resonance. Write down the results in Table 5.1. Table 5.1

Resonance frequencies Resonance number k 1 2 … 10

f1 , кГц

Air

f 2 , кГц

65

f 3 , кГц

 f , кГц

5.3.2.4. To use formula (5.16) for calculations of the sound velocity  , we must use the data from Table 5.1. Then, using formula (5.8) we can calculate  and    . Write down the results calculation in Table 5.2. 5.3.2.5. Estimate absolute and relative measurement error by the method of finding errors of direct measurements. Present the final result in the form:      . 5.3.2.6. Put the results on the graph, the horizontal axis is the resonance number k, and the vertical axis is the difference between the frequency of subsequent resonance and the frequency of the first resonance fk +1 - f1. Draw the best straight line through these points. The slope of the line determines the value of  / 2l (see Eq. (5.17)). Determine the coefficient f / k from the graph, use formula (5.15) to calculate the mean value of the speed of sound. Calculation of the sound speed ki  1

f i  f1 , Гц

i , м / с

i



Table 5.2

and   

 i

 i2

2-1 3-1 . . 10-1

5.3.2.5. Substituting the obtained value of the speed of sound in formula (5.8), calculate the value of   C p / CV and compare it with    from table 5.2. Questions for self-assessment 1. What is called an ideal gas? Write down the equation of state of an ideal gas, and explain it. 2. Formulate the first law of thermodynamics in general and for each isoprocess. Draw the graphs of isoprocesses in coordinates  pV  ,  pT  , VT  . 3. Define the heat capacity of a substance, specific heat and molar heat capacity. Is heat capacity a function of process or state? 4. Why is heat capacity at constant pressure greater than heat capacity at constant volume? What is the physical meaning of the universal gas constant R?

66

5. What is called the number of degrees of freedom? 6. What is the cause of the discrepancy between theoretical and experimental values of heat capacity? 7. What value should γ have for air? References 1. Matveev A.N. Molecular Physics: a Textbook for physical faculties of universities. – 2 nd ed., Revised and ext. / A.N. Matveev. – M.: Vysshaya shkola, 1987. – P. 360. 2. Kikoin A.K., Kikoin I.K.. Molecular physics. – 2nd ed. / A.K. Kikoin, I.K. Kikoin. – M.: Nauka, 1976. – P. 480. 3. Savelyev I.V. The course of general physics. Bk. 3: Molecular Physics and Thermodynamics / I.V. Savelyev. – M.: Astrel. AST, 2003. – P. 208. 4. Laboratory works in physics: textbook / L.L. Goldin, F.F. Igoshin, S.M. Kozel et al..; Ed. Goldin L.L. – M.: Nauka. Main publishing house of Physical and Mathematical Literature, 1983. – P. 704. 5. Zaidel A.N. Errors in measurements of physical quantities / A.N. Zaidel. – SPb.: Lan, 2005. – P. 106.

67

Laboratory work →6 DETERMINAITION OF SPECIFIC HEAT OF LIQUIDS BY COLORIMETRIC METHOD 6.1. The purpose of work: determination of specific heat capacity of liquid by a graphic registration of heat loss due to heat exchange with the environment.

6.2. A brief theoretical introduction Heat capacity Cbody of a body is the ratio of an infinitesimal amount of heat Q obtained by the body to the corresponding increment of its temperature dT: Cbody  Q dT ,

(6.1)

where  Q is the amount of heat needed to change the body temperature by dT degrees. The value of Cbody is measured in J / K. The heat capacity of the body can be referred to the unit of mass, one mole and a unit volume of the body. Therefore, we introduce definitions of specific, molar or volume heat capacity. The heat capacity per unit mass of a substance is called the specific heat с , i.e. Cbody Q . (6.2) с



m

m  dT

Specific heat capacity is measured in J/(kg·K). Heat capacity, referred to one mole of a substance is called the molar heat capacity C: C

Cbody





68

Q .

  dT

(6.3)

Here   M m is the number of moles of the substance, M is its molar mass. The unit of molar heat capacity in SI is J/(molK). It is obvious that C  Mc .

(6.4)

Both specific and molar heat characterize not the body but the material from which the body is composed. Liquid and solid bodies are usually characterized by their specific heat c . On the basis of the first law of thermodynamics, the amount of heat Q received by the system is spent on the increment of its internal energy dU and work done by the system A : Q = dU + A.

(6.5)

As is known, in case of gas heating the work done by it may be different depending on the method of supplying heat. Therefore, the heat capacity of the gas is also a function of the process and may have different meanings depending on the conditions of supply. In particular, we can consider the gas heat capacity at constant pressure C p and at constant volume CV . However, as a result of liquid (as well as solid) heating, its volume increases by a very small value, so the work produced by the liquid during thermal expansion can be neglected compared with the increment of its internal energy (if it is far from the phase transition). Therefore, for the liquid we can assume that C p  CV  C . We may assume that the specific heat of liquids does not depend on the method of heat supply. Specific heat of most liquids slightly increases with increasing temperature. The specific heat of water decreases from 4.2174 kJ / (kgK) at 0 °C to 4.1780 kJ / (kgK) at 35 °C, and then rises to 4.2145 kJ / (kgK) at 99 °C. As it is seen, the maximum relative change in the specific heat of water in the temperature interval from 0 °C to 35 °C does not exceed 0.9%. Therefore, for most practical calculations we can ignore the dependence of specific heat on temperature and take for the specific 69

heat of water its mean value, c = 4.1868 kJ / (kgK) in the temperature interval from 0 °C to 100 °C. In most methods of calorimetric measurements, distilled water is used as a reference body with known specific heat. In this paper, the specific heat of the studied liquid is determined by the graphic account of heat loss due to heat exchange with the environment. In the experimental determination of the specific heat by formula (6.2) there always arise problems related to the fact that during heating of the investigated body part of heat is released to the environment through heat exchange. Therefore it is always necessary to make corrections for the loss of heat to the environment. The method of account for corrections depends on the colorimetric method used for determining the specific heat capacity of the body. In this work the method of calorimetric measurements of the specific heat capacity of the body is used, when the calorimeter temperature is kept above the ambient temperature. Hence, throughout the experiment the calorimeter only gives off heat to the environment. In this experiment the electric heater, dipped into the calorimeter with the tested liquid, is switched on and the liquid is heated to a temperature of 10 ÷ 15 °C above the ambient temperature. Then the heater is switched off, and the system cools and gives off heat to the environment by the action of natural convection. During the experiment students observe changes in the temperature of the liquid in the calorimeter and the ambient temperature, and write these values down in the table. For the case of switched-on heater on the basis of the heat balance equations we can write: I  U  dT  cm  c0 m0  dT   Tsur  Tenv S d

(6.6)

For the case of switched-off heater:

cm  c0 m0  dT   Tsur  Tenv S d  0 .

(6.7)

Here I, U are current and voltage on the heater; c , m are specific heat capacity and mass of the tested liquid in the calorimeter; 70

c 0 and m0 are heat capacity and the mass of the calorimeter;

 is the coefficient of heat transfer from the calorimeter to the environment (see lab work No. 4); S is the surface area of the calorimeter; Tsur and Tenv are the mean temperatures of the calorimeter and the surrounding surface of the medium; dT is an increment of the average over the volume temperature of the liquid during time d . As a rule, all dielectric liquids have a low coefficient of thermal conductivity and in the absence of convective mixing the distribution of the liquid temperature throughout the volume may be non-uniform. The liquid temperature is not equal to the surface temperature of the calorimeter wall. It is therefore necessary during the experiment to continuously mix the fluid in the calorimeter with a stirrer. Only under this condition will equations (6.6) and (6.7) be valid. It can also be assumed that the surface temperature of the calorimeter Tsur is equal to the average temperature of the liquid T: Tsur  T . The heat transfer coefficient  weakly depends on the temperature difference Tsur  Tenv , so it can be considered a constant quantity independent of temperature. If the heater is switched on at the moment of time  1 , when the calorimeter with liquid has an average temperature T1 , and is switched off at the moment of time  2 at the average temperature T2 , the heat balance equation (6.6) can be written in finite differences: T T  IU ( 2   1 )  (cm  c0 m0 )T2  T1     1 2  Tcp S  2   1  .  2 

(6.8)

In the last term the substitution Tsur  T1  T2 was made as the heat 2

removal to the environment is determined by the average tempera71

ture difference between the heated body and the environment during the experiment: (T  Tenv )  (Tsur2  Tenv ) T1  T2 (6.9) Tsur  Tenv  sur1   Tenv 2 2 taking into account the assumption of the equality of the mean temperature of the liquid and the surface of the wall of the calorimeter. To determine the heat transfer coefficient  we use the equation (6.7) for the heat balance for switched-on heater, which can be rewritten as: S dT (6.10) d .  Tsur  Tenv

cm  c 0 m0

Taking into account that Tsur  T , (6.10) can be written as: S dT d .  T  Tenv cm  c 0 m0

(6.11)

Let us integrate (6.11) over time starting from the moment of turning off the heater  2 to a moment  , when the calorimeter cools from T2 to T . ln

Then

T  Tenv S  2    .  T2  Tenv cm  c 0 m0

T  Tenv  (T2  Tenv ) e

Where the coefficient k

 k (  2 )

S cm  c0 m0

,

(6.12)

(6.13)

is called the cooling rate. This shows that the calorimeter temperature during cooling decreases with time exponentially. From the experiment, we can easily determine the value of k plotting the dependence of ln T  Tenv  on  . Substituting  S from (6.13) into (6.8), we obtain the final formula for determining the specific heat capacity of the tested liquid: 72

c

IU ( 2   1 )    T  T2  m T2  T1    1  Tenv ( 2   1 )k   2   



m0 c 0 . m

(6.14)

6.3. Description of the experimental setup The calorimeter is (see Figure 6) an aluminum glass 3 with a tested liquid and an electric heater 8 in it. The calorimeter is attached to the lid and is set inside a large cylinder 2 with double walls, which plays the role of the thermostat. During the experiment through the walls of the thermostat water from the water tap is supplied and a constant temperature of the medium Tñð is maintained. The electric heater is connected to the autotransformer through additional resistor R. The electric current passing through the heater is measured by an ammeter, the voltage drop is measured by an AC voltmeter. The values of the liquid temperature in the calorimeter and water in the thermostat are measured by a mercury thermometer with an accuracy of 0.1 K

1 – the control unit, 2 – thermostat, 3 – calorimeter, 4 – flanges, 5 – lid, 6 – thermometer, 7 – chromel-copel thermocouple, 8, 9 – stirrer, 10 – cooling jacket, 11 – cylindrical slot for the thermometer, 12 – terminals, 13 – handle mixer, 14 – sleeve, 15 – heater power cable Fig. 6. Setup for measuring the specific heat of liquid by the method of periodic heating and cooling

73

6.4. Procedure of work fulfillment 6.4.1. Turn on the cold water passing through the thermostat and wait until the temperature of the environment Tñð stops changing. 6.4.2. Record in the workbook specifications of the setup:  calorimeter mass m0 ;  calorimeter mass with liquid m  m0 ;  mass of the tested liquid m;  specific heat capacity of the calorimeter c0 . 6.4.3. Turn on the heater by passing a current of about 1.0 ÷ 1.5 A through it. When the temperature of the liquid in the calorimeter is 10 ÷ 15 °C higher than the coolant temperature in the thermostat, switch the heater off the power. 6.4.4. The first period. Turn on the stopwatch and every minute record in Table 6.1 the liquid temperature T in the calorimeter and the temperature Tenv of water flowing in the thermostat by the readings of the corresponding thermometers. It is necessary to continuously mix the liquid in the calorimeter manually rotating the stirrer with the impeller at the end. Do measurements for 15 minutes. This will be the first period of the experiment. Do not stop the stopwatch till the end of work. Measurement of the temperature of the environment Tenv and the temperature and the tested liquid T τ, min

Tenv, °С T, °С First period

τ, min Tenv, °С T, °С Second period

τ, min

Table 6.1

Tenv, °С T, °С Third period

6.4.5. Second period. In 15 min turn on the heater by passing a current of about 1.0 ÷ 1.5 A through it. Write down the values of current and voltage on the heater. Record the thermometer readings every 0.5 minutes during the second 15-min period. 6.4.6. The third period. In 15 minutes, turn off the heater and continue recording thermometer readings every minute for 15 minutes. 6.4.7. Turn on the stopwatch. The experiment consists of three periods, 15 minutes each. 74

6.4.8. By the results of experiments determine the difference T  Tenv and the natural logarithm ln(T  Tenv ) for each measurement. Write them down in Table 6.2. Plot the graph of the dependence of ln(T  Tenv ) on  during the experiment. Data for plotting the graph τ, min 0 1 … 15 15,5 16,0 … 30,0 31 32 … 45

T – Tenv, °С First period

Table 6.2

ln(T – Tenv)

Second period

Third period

6.4.9. By the tangents of slope angles of the parts of the graph corresponding to the cooling mode before and after the heating section (the first and third periods), determine the rates of cooling k1 and k  k2 . k 2 , and the average value of k  1 2 6.4.10. Substitute the value of k in the formula (6.14) and find the specific heat of the studied liquid. 6.4.11. Compare the result obtained in the experiment with the table values of specific heat for the tested liquid and explain possible reasons for differences between them. Questions for self-assessment 1. What is the heat capacity and what does it describe? 2. What assumptions were made in the derivation of formulas (12) and (14)? 3. How are heat losses of the calorimeter to the environment taken into account in this work? 4. Why does the graph of time dependence of water temperature in the calorimeter in semi-logarithmic coordinates consist of segments of straight lines?

75

References 1. Physical practicum. Mechanics and Molecular Physics. Ed. Iveronova V.I. – M.: Nauka, 1967. – 352 p. 2. Matveev A.N. Molecular physics. – M.: Vysshaya shkola, 1987. – 360 p. 3. Sivukhin D.V. The general course of physics. V. 2. Thermodynamics and molecular physics. – M.: Nauka, 1979. 4. Tables of physical quantities. Reference book. Ed. Kikoin I.K. – M.: Atomizdat, 1976. – P. 1006. Task for SRW 1. Measure the mass of the empty calorimeter. Pour distilled water into the calorimeter, measure the mass of the calorimeter with distilled water. Find the mass of water. Assemble the setup. 2. Repeat the task of pp 6.4.1-6.4.10 for distilled water and find the experimental value of its specific heat. 3. Compare the experimental value of specific heat capacity of water with table values and estimate the accuracy of measurements on the setup. Estimate possible reasons for these differences and ways to overcome them.

76

Laboratory work →7 DETERMINATION OF AIR VISCOSITY 7.1. The purpose of work: experimental determination of air viscosity.

7.2. Theoretical introduction 7.2.1. Viscous friction. Let us consider a liquid layer enclosed between two plates located at a distance y from each other (Figure 7.1).

Fig. 7.1. Definition of viscosity

Suppose that the lower plate does not move, while the upper plate of area S moves at a constant speed 0 in the direction of axis x   under the action of an external force F . The friction force f applied  to the upper plate is equal and opposite to the external force F (Figure 7.1 a). Due to the internal friction of the liquid layers located very close to the surface of the body move nearly at the speed of movement of the body surface area. If y  S , the speed of liquid between the plates will increase linearly from 0 to  0 when the vertical coordinate of the liquid layer changes from 0 to y (Fig. 7.1.b). 77

Newton suggested the following equation for the force of viscous friction:  (7.1) fx  0 S y

where the proportionality coefficient  is called liquid (or gas) viscosity and depends only on the properties of liquid (gas) filling the space between the surfaces (at a given temperature). The ratio of the friction force f x to the area of the plate surface S

 xy  

0 y

(7.2)

is called friction-induced shear stress. Here, indices x and y indicate the direction of movement (x) and the direction in which the velocity changes (y). Imagine that the liquid between the surfaces is divided into thin parallel layers. Each layer is moving uniformly, the upper layer pulls forward the layer lying under it with a force equal f, whereas the lower layer pulls the adjacent upper layer with a force equal to f and directed backward. Thus, the friction force f is transmitted from one liquid layer to another, hence, from one surface to another. Two equal in magnitude and opposite in direction forces act on each layer, therefore its movement is uniform. The above-stated fully relates to the friction shear  xy , as it is equal to the friction force acting on a unit of surface area. In SI viscosity  is measured in Pas (Pascalsecond). Viscosity  is different for different gases and increases with increasing temperature. For gases, viscosity  is related to the mean free path  l  by the following equation: 

1  l  u   , 3

where  u  is the arithmetic average velocity of molecules;  is the gas density at a given temperature. 78

(7.3)

It is known that for an ideal gas  u 

pM , 8RT ,  RT M

(7.4)

where M is the molar mass of the gas; p is pressure; T is the absolute temperature. From (7.3) and (7.4) we obtain:  l 

3  RT . p 8M

(7.5)

Equation (7.5) enables us to determine the mean free path of gas molecules by the experimental values of viscosity, temperature and pressure. Using the formula from the elementary kinetic theory of gases  l 

1 kT ,  2 2 d n 2 d 2 p

(7.6)

we can determine the effective diameter of gas molecules. The other quantity used in calculations along with the dynamic viscosity  is the kinematic viscosity v of liquid (or gas), defined as:

   , where  is the density of the liquid (or gas). Kinematic viscosity is measured in SI in m2 / s. In general, when near the surface of the body the fluid velocity changes arbitrarily in the direction perpendicular to the surface section under consideration, the numerical value of the tangential friction stress determined by the formula:

 xy   79

d x , dy

(7.7)

where у is the direction of normal to the surface element, d x – the derivative of the velocity projection on the x-axis by y dy

coordinate. 7.2.2. The flow of viscous fluid in the tube. Let us consider a steady laminar flow of a viscous fluid in a circular tube and find the law of variation of the flow velocity  as a function of distance r from the axis of the tube. Let us choose within the tube of radius R an imaginary cylindrical volume of radius r and length l (Figure 7.2).

Fig. 7.2. To the derivation of the dependence of the flow rate on the distance from the axis of the tube

In case of steady flow in a tube of constant cross section, the flow velocities of all the particles remain unchanged. Therefore, the sum of all external forces acting on the selected volume of fluid is equal to zero. In the flow direction along the x axis acts the force caused by the pressure difference in sections 1 and 2: Fpress  ( p1  p2 )   r 2 .

The friction force acts on the side surface of the selected volume of liquid: F fric   xr  2 r l . The sum of projections of these forces on the flow direction must be zero: (7.8) Fpress  F fric  ( p1  p2 )   r 2   xr  2 r l  0 . 80

Hence, we get:

p1  p 2 r  . l 2

 xy 

From this equation it is seen that the tangential friction stress changes along the section of the tube linearly increasing from zero at the axis of the tube r  0 to a maximum value  max 

p1  p2 R  l 2

on the wall of the tube r  R . Substituting into formula (7.8) the expression for the frictional stress (7.7) d x ,  xr   dr we obtain a differential equation for the flow speed: d x ( p  p 2 ) r .  1 dr 2 l

(7.9)

Integrating the last equation, we find:

 x (r )  

 p1  p2   r 2 4l

C.

(7.10)

The integration constant C is found from the boundary conditions of adhesion of viscous fluid to the walls of the tube, i.e.  x  0 for r  R (R is the tube radius): C

p1  p2 R 2 .  l 4

Substituting (7.11) into (7.10), we get: 81

(7.11)

 x (r ) 

p1  p 2 1 2  R  r2 . l 4





(7.12)

From this expression it is seen that the speed is distributed over the tube cross-section according to the parabolic law. Its magnitude is equal to zero at the tube wall r  R and reaches a maximum value  0 at r  0 :

0 

p1  p 2 R 2 .  l 4

(7.13)

Taking into account (7.13), we transform the formula (7.12) to the form: (7.14)  (r )  0 1  r 2 / R 2  The profile of flow rate is shown in Figure 7.2. Knowing the velocity distribution over the cross section of the tube, it is possible to calculate volumetric flow rate per second q through the tube cross-section. To do this, choose in the cross section of the tube an elementary ring of radius r and thickness dr so small that inside this ring the speed of fluid flow can be assumed to be constant (Figure 7.3). Then, through the cross-section of a thin ring of 2rdr in area the liquid volume (7.15) dq   2 rdr. will pass per second. Substituting the value of the velocity from formula (7.14) and integrating over the entire cross section of the tube, we find the volume flow rate per second: R





q  2  0 1  r 2 / R 2 rdr  0

p1  p2  R 4 .  l 8

Finally, we obtain the Poiseuille law: q

p1  p2  R 4 .  l 8 82

(7.16)

Fig. 7.3. To the derivation of the Poiseuille formula

This law is used for experimental determination of viscosity  by the measured values of volumetric flow rate per second q, radius of the tube R and pressure drop p1  p2 on the tube section of length l:  R 4  p1  p2  (7.17)  8ql Knowing volumetric flow rate per second, we can find the average flow speed over the cross section of the tube:  

q p1  p2 R 2 .    R2 l 8

(7.18)

Comparing (7.13) and (7.18) we see that the maximum flow speed at the axis of the tube  0 is 2 times greater than the average flow rate over the cross-section:

0  2    . 7.3. Description of the setup and experimental technique 7.3.1. Accessories: a capillary tube (a cylindrical tube), fixed on the vertical rack, a micromanometer of MCM type, a gas meter, a stopwatch, a volumetric flask. 7.3.2. The experimental setup is shown in Figure 7.4. The air flow through the capillary 5 is created by the pressure drop at its ends. The 83

lower end of the capillary through the tee-joint 11 is connected to the gas meter 1 filled with water. When the water flows out from the gas meter through the valve 3, a negative pressure is formed above the surface. Hence, the pressure at the lower end of the capillary is less than the atmospheric pressure. The pressure drop at the ends of the capillary is measured by the micromanometer 8 connected to the gas meter and the lower end of the capillary through the tee-joint 11.

1 – tank, 2 – an inlet valve, 3 – a drain valve, 4 – a tripod, 5 – capillaries, 6 – a hose, 7 – a hose, 8 – a micromanometer, 9 – a fitting, 10 – a connector, 11 – a tee-joint glass valve, 12 – a tee-joint, 13 – a level meter Fig. 7.4. Setup for determining the viscosity of air flowing through a capillary as a laminar flow

The scale of micromanometer 8 graduated in millimeters of water column, enables us to measure the pressure difference up to 250 mm H2O (the working fluid in the micromanometer is ethyl alcohol). To increase the sensitivity of micromanometer the measuring tube is inclined. The difference between the atmospheric pressure and the pressure at the lower end of the capillary in Pa is determined by the length n of the alcohol column in divisions on the micromanometer scale multiplied by the coefficient k (0.2; 0.4; 0.6; 0.8;) corresponding to the inclination of the measuring tube, and the acceleration of free fall g. Alcohol meniscus is fixed at zero in the measuring tube of the manometer by means of the cylinder, the depth of immersion in which in the alcohol is regulated by the screw. The micromanometer has two levels, perpendicular to each other on the plate. The device is fixed on a certain level with the help of leveling feet. 84

A 3-way valve having two operating positions is mounted on the cap of the micromanometer. In the first operating position, the difference between atmospheric pressure and pressure in the system is measured: through the junction (+) – if the pressure in the system is higher than the atmospheric pressure, via the junction (-) – if the system pressure is lower than the atmospheric one. In the second position, the reservoir of micromanometer and the upper end of the measuring tube are connected to the atmospheric pressure, in this position zero setting of the scale is made. Volumetric flask is designed to determine the volume of liquid flowing from the gas meter over a time  determined by the stopwatch. 7.4. Procedure of work fulfillment 7.4.1. Study all the devices used in the work. Write down in Table 7.1 data on the length of the capillary l, the capillary radius R at ambient temperature t and barometric pressure patm. 7.4.2. If the gas meter is not filled with water up to the upper level (watch the liquid level in the water-metering tube), ask the laboratory assistant to fill the gas meter with water. 7.4.3. Connect the micromanometer to the tee-joint through the junction (-) and set the micromanometer by levels. Place the measuring tube of the micromanometer on the slope specified by the teacher. Check that the device is set on zero. 7.4.4. Slowly opening the valve of the gas meter (Figure 7.4), choose such water velocity that the pressure drop at the ends of the capillary is within the upper half of the scale. 7.4.5. After getting a stationary water flow (micromanometer readings do not change), using the measuring flask of volume V and the stopwatch determine the time during which the given water volume flows out from the gas meter. Write down the experimental data in Table 7.1. The results of measurements of flow rate and pressure drop at the ends of the capillary l=…m, t =… °С p àòì = … Pa, R =… m, № k V, m3 , div n ni , div nk , div 1. 2. 3.

85

 ,c

Table 7.1

q ,m3/c

7.4.6. If during the experiment 7.4.5. the micromanometer readings change significantly write down the readings at the beginning ( n í ) and end ( n ê ) of the experiment. Repeat the experiment 3 times. 7.4.7. Gradually reducing the speed of water flow, so that the readings of micromanometer decreased each time by 1.5 times, determine the volumetric flow rate for different pressure drops at the ends of the capillary. The experiment is finished at the lowest pressure drop of about 20 divisions on the inclined scale of the micromanometer. All experiments are conducted in the order specified in paragraphs. 7.4.5. and 7.4.6. The obtained results are written in Table 7.1. 7.4.8. By the data of each measurement determine the volumetric flow rate of liquid per second q  V , the average value  nн  n к , and write them down in Table 7.1.  n  2

7.4.9. The pressure drop in Pa at the ends of the capillary is defined by the formula: p  p1  p2  9,8 k  n  ,

where  n  is expressed in divisions, k is the coefficient of micromanometer inclination. Therefore, the working formula for determining the dynamic viscosity of air  is:

  3.848

k  R4  n  . q l

(7.19)

Substituting k , q , R ,  n  , l in formula (7.19), calculate the value of air viscosity  for each experiment and record it in Table 7.2. Write down in the table the values of the average speed over the cross section    for each current mode, which are calculated by the following formula:

   86

q . R 2

Table 7.2 The results of calculation of the flow rate q, the coefficient of viscosity  and the average speed over the cross section q, m3/s

p ,

,

  ,

Pa

Pas

Pas

 , Pas

 2 , Pa2s2

  , m/s

7.4.10. Determine the average value of  by the results of all the experiments and estimate the error by the method of direct measurements. 7.4.11. Using maximum and minimum values of speed and the volumetric flow rate per second, determine maximum and minimum values of the Reynolds number: Re 

2R    





2 q ,  R

(7.20)

where  is the air density. The air density is determined by the formula:

  0

p atm T0 ,  p0 T

(7.21)

where  0 = 1.2928 kg/m3 is the density of dry air at atmospheric pressure p 0 , p 0 =1.013105 Pa, Т0 = 273К, p atm and T  273  t °C are pressure and temperature at which the experiments were conducted. Air density at the temperature and pressure of the experiment can also be found using the Mendeleev-Clapeyron equation. 7.4.12. By the maximum value of the Reynolds number estimate in which mode of the air flow in the capillary the experiments were made. If Re  2300 , the flow is turbulent and the Poiseuille formula is not valid. If Re  2300 , the flow is laminar and the Poiseuille formula can be applied. 87

7.4.13. Compare the value  obtained in your experiments with the table value of viscosity for dry air, which can be found by the Sezerlend formula, which for air is written as: 3

  1,528  10

6

273  t 2

(Pas),

403,6  t

(7.22)

where t is the room temperature in °С. 7.4.14. Using the value of viscosity  found from the experiments and formula (7.5), find the mean free path  l  . 7.4.15. Knowing the mean free path  l  , estimate the effective diameter of the nitrogen (oxygen) molecule from formula (7.6): d2 

kT 2  l  p

.

Compare the obtained with the table value. 7.4.16. Try to explain the reasons for discrepancies between the results of the experiment and the table value and formulate your conclusions. Questions for self-assessment 1. Formulate Newton's law of viscous friction. Define viscosity. In what units is viscosity measured in the SI? 2. What are the units of pressure in the SI? What non-SI units of pressure do you know? 3. Derive the Poiseuille formula for a steady motion of a viscous fluid in a cylindrical tube. 4. What is called the mean free path of molecules? How can we find this value? 5. Explain the conditions of fluid flow in a tube, in which the Poiseuille calculations are valid. 6. Explain the principle of operation of the laboratory setup and micromanometer. References 1. Matveev A.N. Molecular Physics: a Textbook for physical faculties of universities. – 2 nd ed., Revised and ext. / A.N. Matveev. – M.: Vysshaya shkola, 1987. – P. 360.

88

2. Kikoin A.K., Kikoin I.K. Molecular physics. – 2nd ed. / A.K. Kikoin, I.K. Kikoin. – M.: Nauka, 1976. – P. 480. 3. Savelyev I.V. The course of general physics. Bk. 3: Molecular Physics and Thermodynamics / Savelyev I.V. – M.: Astrel. AST, 2003. – P. 208. 4. Laboratory works in physics: textbook / L.L. Goldin, F.F. Igoshin, S.M. Kozel et al..; Ed. Goldin L.L. – M.: Nauka. Main publishing house of Physical and Mathematical Literature, 1983. – P. 704. 5. Tables of physical quantities. Reference book / I.K. Kikoin, I.S. Grigoriev, Yu. Danilov et al.; Ed. I.S. Kikoin. – M.: Atomizdat, 1976 – P. 1006. Task for Student research work 1. Purpose of SRW: 2. Determine a more precise value of air viscosity, taking into account corrections for pressure drop at the inlet of the capillary and in the accelerated section of the flow, where the Poiseuille formulas are invalid; 3. Determine the dependence of the hydraulic resistance of a cylindrical tube coefficient on the Reynolds number; 4. Determine the critical Reynolds number corresponding to the transition from laminar to turbulent flow in the fluid motion in a cylindrical tube. 5. Calculation of corrections to the Poiseuille formula and derivation of formulas for calculation of the hydraulic resistance of the tube. 6. Laminar movement in the accelerating section of the tube. In the derivation of the Poiseuille formula (7.16) we assumed that the velocity distribution in all sections of the tube is the same (i.e. a steady flow). For tubes of finite length it is valid only for parts located at a great distance from the point of entrance into the tube. The experiment shows that if the fluid enters the tube from the reservoir, the dimensions of which are quite large compared to the diameter of the tube, and the entrance into the tube is rounded so as to avoid flow disturbances, then in the input section the speed in all points is constant and equal to the average flow fluid speed (except for a very thin layer near the tube wall). As the liquid moves along the tube, the liquid layers adjacent to the walls are decelerated by the action of viscous forces. The thickness of the boundary layer grows along the length of the tube until it fills all its cross section. As the distance from the entrance increases, the velocity profiles change gradually approaching the parabolic shape of a steady laminar flow (Fig.7.5). The part of the tube from the inlet section to the section, where the speed at the tube axis differs by 1% from the maximum speed of the parabolic profile, is called the initial or accelerating part. The length of the initial sector



is determined by a semi-empirical formula:

xí  0,04 d Re ,

(7.23)

where d – is the tube diameter. Thus, the Poiseuille (7.16) formula is valid for the tube sections located downstream the accelerating part of the flow. However, in most cases it is possible

89

to experimentally measure only the pressure drop along all length of the thin tube, starting from the inlet section.

Initial section

A Steady laminar flow

Fig. 7.5 In this case it is necessary to take into account the correction for the pressure drop in the first section. If the gas or liquid flows into the tube from the reservoir, the diameter of which is greater than that of the tube, then applying the Bernoulli equation to the cross section of the reservoir far from the entrance to the tube and at the entrance section of the tube, we get:    2 , (7.24) p0  p1  2 where p 0 is the pressure in the reservoir;

p1 is the pressure in the inlet section of the tube. In the accelerating section of the tube, with an increase in the thickness of the boundary layer the speed in the axial part of the flow increases from    to max  2    at the end of the section. The kinetic energy of the flow passing per unit time through the cross section of the tube increases from its initial value:

E0 

m    2   R 2   3 ,  2 2

(7.25)

where m    R 2    is the mass of liquid flowing into the tube per unit of time, to the value E at the end of the acceleration section, where a parabolic velocity profile is formed: R

E  2   0

 3r 2

dr    R 2    3 .

(7.26)

Thus, the kinetic energy E of the flow with a parabolic profile is 2 times greater than the initial kinetic energy E0. To provide this additional kinetic energy to the flow, an additional pressure drop

90

   2 . 2

(7.27)

occurs in the accelerating section of the tube. A higher velocity gradient near the tube walls in the accelerating section creates a pressure drop greater than the pressure drop obtained from the Poiseuille law for the same length, by the value

0,41

   2 . 2

(7.28)

Summing up all kinds of additional losses in the accelerating phase (7.24), (7.27) and (7.28) and adding them to the pressure loss along the tube length from the Poiseuille formula (7.16), we obtain

p0  p2 

8  l        2 .  2,41 2 R 2

(7.29)

Here p 0 and p 2 are static pressures in the reservoir before the inlet section and

in the section at a distance l from the inlet of the tube. Substituting the volumetric flow rate of fluid per second instead of    and taking into account that p0  p2  9,8 k n , from (7.29) we can find the value of viscosity:



9,8k n R 4 2,41 q q. knR4   3,848  0,048 8ql 16 l ql l

(7.30)

This formula is used to determine air viscosity for laminar flow in the tube with account for the influence of the accelerating section. 7.7.2.2. Calculation of hydraulic tube resistance. When the liquid moves in the tube of constant cross-section of a diameter d , the pressure loss p1  p 2 on the frictional resistance for the section l is determined by the empirical Darcy law:

p1  p2   

l    2 .  2 d

(7.31)

Here λ is the coefficient of hydraulic resistance of the tube. If the motion of fluid in a cylindrical tube is laminar and steady, then the part of the tube located far from the entrance the pressure difference p1  p 2 is determined by the Poiseuille formula (7.16). Substituting (7.16) into (7.31), we find the coefficient of hydraulic resistance:



where Re 

p1  p 2 64 ,  l      2 Re  2 d

 d    is called the Reynolds number. 

91

(7.32)

Formula (7.32) is called the Poiseuille law for the coefficient of hydraulic resistance in the laminar flow. The Reynolds number plays an important role in the similarity theory of hydrodynamic flows. It is experimentally established that laminar or layered flow of viscous fluid with a parabolic velocity profile in the pipe forms only at Reynolds numbers Re  2300 . At Reynolds numbers Re  2300 the flow fundamentally changes its character: a vortex or a turbulent flow appears instead of a quiet layered flow. In the transition from the laminar to the turbulent flow the coefficient of hydraulic resistance increases rapidly and in the region of a developed turbulent mode the dependence of λ on Re number obeys empirical Blasius’s law:



0,3164 , Re1/ 4

(7.33)

valid for values: 2300  Re  10 5 . As in our experiments we determine the pressure drop along the length of the tube, from its inlet section, to determine λ for the laminar flow it is necessary to take into account the correction for the pressure drop in the accelerating phase, using the formula (7.29). Thus, the formula for determining λ for the laminar flow from the experiment data is written as:



 p0  p2   2,41       l      d 2

2

2

2

 3,87

k  n  R5 R  4,82 . l   l  q2

(7.34)

7.8. Procedure of work fulfillment 7.8.1. Do the work in accordance with paragraphs 7.4.1 – 7.4.3. Place the measuring tube of the micromanometer put in a position corresponding to the slope coefficient k = 0.8 or, use a U-shaped pressure gauge for large pressure drops. 7.8.2. Slowly opening the valve of the gas meter, get the maximum flow rate per second at which the micromanometer reading is close to the upper limit of the scale. Measure the time  during which the liquid volume V, corresponding to the volume of the volumetric flask, flows out and write it down in Table 7.3. 7.8.3. Set the water flow so that each subsequent experiment the level of liquid in the micromanometer tube decreased by 20-40 divisions and repeat the measurements to the values of n = 20-30 divisions for the slope coefficient k = 0.8. 7.8.4. After this move the micromanometer to the slope k = 0.2 and repeat measurements from values n = 240 divisions to n = 20-30 divisions similar to paragraph 7.8.3. 92

7.8.5. Calculate the coefficient of hydraulic resistance  by the formula (7.34) for each experiment and write down in Table 7.3. Calculation of hydraulic drag coefficient № пп 1. 2. 3.

k

n, дел.

V, м3

, с

q, м3/с

, кг/мс

Re



lg 100

Table 7.3

lg Re

7.8.6. Using the experimental data calculate the viscosity of air

 by the formula (7.30) for several values of volumetric flow rate

per second. Find their mean value and the error by the method of direct measurements. Compare the value of  with the result obtained from the formula (7.17) without corrections and with the table value of air viscosity. 7.8.7. Using the experimental value of viscosity  and the value of air density  calculated from the formula (7.21) calculate the Reynolds number Re  2  q for each measurement and write in  R Table 7.3. 7.8.8. Find logarithms of Re , 100  values for each test and write them down in Table 7.3. Before taking the logarithm is necessary to multiply by 100 to get positive numbers. 7.8.9. Using the experimental data make a graph of lg(100  ) on lg Re . Put the experimental data as points on the graph. On the same graph draw a theoretical dependence of lg(100  ) on lg Re , using the formula (7.32). If we multiply both sides of this equation by 100 and take logarithms, we get:

lg(100 )  lg 6400  lg Re .

(7.35)

It is seen that in logarithmic coordinates the graph of this dependence is a straight line. To construct the line it is sufficient to find two of its points corresponding to values Re  200 and Re  2000 and to draw a straight line passing through those points. 93

If measurements are made with sufficient accuracy, the experimental points will lie close to the theoretical straight line. Near the critical value Re  2300 , the experimental data can be very different from the Poiseuille law (7.32) but close to the Blasius law (7.33). In the area of sharp deviation of the experimental data from the Poiseuille law we can see a transformation of the flow in the pipe from laminar to turbulent motion. By the experimental data from the graph find the critical Reynolds number, if you have reached the turbulent flow regime. 7.8.10. Analyze the results and write your conclusions. References 1. Fabrikant N.Ya. Aerodynamics. – M.: Nauka, 1964 ch. – V. I, § 2, 3. – P. 583-590.

94

Laboratory work →8 STUDY OF TEMPERATURE DEPENDENCE OF LIQUID VISCOSITY 8.1. The purpose of work: 8.1.1. Determine the dependence of fluid viscosity on temperature by capillary viscometer using the Poiseuille formula. 8.1.2. Using Ya.I. Frenkel’s model of internal friction in the fluid, interpret the temperature dependence of viscosity and evaluate the activation energy needed to move a molecule from one temporary equilibrium position to another.

8.2. A brief theoretical introduction Development of the theory of the liquid state of matter is one of the most important problems of modern statistical physics. A specific feature of the liquid state is intensive interaction of its particles under the condition of their high disorder. This feature of the liquid state hampers the theoretical analysis of the problem. Therefore, the development of the liquid theory is behind the theory of gas and the crystalline state of matter. A crucial role in the development of modern high-level statistical theory of gases and crystals played the fact that in both cases we can use easily foreseeable limiting cases of the ideal gas and ideal crystal. An ideal gas is a very rarified space of particles with a high degree of randomness and a perfect crystal is a case of perfect ordering of particles under the condition of high density. For the liquid, there is not such a simple and easily visualized model, which could be taken as zero approximation in the construction of liquid theory. More detailed analysis of some of the properties of gases, liquids and solids leads to the conclusion that liquid occupies an intermediate position between a crystal and gas. The first X-ray diffraction studies of liquids led to significant changes in the views on the nature of the liquid state. It was found 95

that liquids are not structureless as gases. Namely, liquids have a certain order in the arrangement of their particles, especially near the solidification point. However, this order exists at small distances of about 2-3 molecular layers. In this case we speak about a short-range order, in contrast to the long-range order – the order at large distances. Thus, the nearest neighbors of each liquid molecule, conventionally called «central», are arranged in a certain order, forming a structure similar to the structure of a solid. However, as the distance r between the molecules increases, the dependence between the location of conditionally selected first «central» molecule and its more distant neighbors weakens. At r  (3  4)d , where d is the effective molecular diameter, such a dependence, in contrast to the crystals, is practically absent. Therefore it is said that liquids have a short-range order, whereas crystals have a long-range order. Liquids occupy an intermediate position between gases and crystals not only by their structure and intensity of intermolecular interaction, but also by the way of their thermal motion. The theory of thermal motion of fluid particles was developed by the Soviet scientist Ya.I. Frenkel. In crystals we imagine the thermal motion of atoms, ions or molecules in the form of vibrations near a strictly-defined position, in rarefied gases it is disorderly collisions of molecules at the end of their free path. In the case of thermal motion in liquid, it is a combination of vibrational and translational motion of its particles. Therefore, for some time each molecule fluctuates around a certain equilibrium position, and then abruptly moves to a new equilibrium position, spaced apart from the previous one by a distance approximately equal to the size of the molecule. Hence, the molecules only move slowly within the fluid, staying for some time near certain places, in the so-called «resident state» Such a behavior of molecules can be explained as follows. In the 1970s, John. D. Bernal put forward the hypothesis that in liquids we encounter the so-called symmetry of the fifth order. This means that in the plane each particle has on average of about 5 nearest neighbors, and in a space package – this number is about 11. From this statement it follows that: 1) It is easy to see that in this type of package only a short-range order can exist, as the plane can be covered with a mesh consisting of 96

triangles, squares, hexagons, but not pentagons. Similarly, the space cannot be filled by polyhedrals whose faces would be regular pentagons. 2) The polyhedrals, characterizing the package of particles in liquid, have slightly different ribs. In crystals, all particles are at a distance d from each other, where the attractive forces are equal to the repulsive forces, in liquids the distance between molecules on some ribs is a little larger than d and on the other ribs it is smaller than d, and on average it is equal to d (about 10-10 m). 3) Polyhedrals with the symmetry of the fifth order, cannot form a perfect structure in the dense package, they form some compact spatial structures called pseudokernels. There must be large spaces between these kernels called holes in Frenkel terminology. 4) Pseudo kernels are not sufficiently stable formations. Due to the large number of «holes» particles can easily move from one kernel to another, and fluid structure changes continuously. Hence, the liquid particle vibrates about the instantaneous equilibrium position for some time – residence time. Then it jumps to the location of the hole and joins the other pseudokernel. The average time of «resident life» – a period of time during which the structural configuration of the pseudokernel characterizing the short-range order does not change – is called the relaxation time  . To jump from one temporary equilibrium position to the other, the liquid molecule must break its connections with the neighboring molecules. After transition to a new equilibrium position, the molecule binds with the new «neighbors». The first process requires energy, which is released when new connections are formed. Hence, when a molecule leaves its old place, its potential energy must increase by U a . This transition is referred to as a transition through the potential barrier of height U a , and U a is called the activation energy. Transition through the potential barrier occurs in case of a random increase in the thermal energy of motion of individual molecules due to the transfer of this energy by the other molecules. Calculations show that the relaxation time  depends on the activation energy U a and the absolute temperature T: Ua kT

  0  e , 97

(8.1)

where  0 is the average period of molecular vibrations about the equilibrium position, k is Boltzmann constant, k = 1.38066210-23 J/K. The relaxation time  depends on temperature. With increasing temperature, the relaxation time decreases rapidly, which leads to a greater mobility of molecules in liquid at high temperatures. We can estimate the order of magnitude of the period of one vibration  0 . If we put in equation (8.1) U a  0 , the time    0 must be of such order of magnitude that a molecule moving at its average thermal velocity  x in the direction x could pass during this time the path equal to   1010 m . Then:

 0   /x .

(8.2)

The speed of random thermal motion of molecules in a liquid can be determined by using the relationship between the kinetic energy of the thermal motion of molecules and the temperature: m x2 / 2  kT / 2 ,

hence,  x 

kT m ,

(8.3)

where m is the mass of one molecule. At room temperature, the average thermal velocity  x has the order of hundreds of meters per second. Therefore, assuming that  x  0,5  103 m / s ,   10 10 m , we obtain  0  2  10 13 s . The relaxation time τ can be very different, but    0 . For lowviscous liquids (liquefied noble gases)   1011c , whereas for resins it is up to several hours or even days. The viscosity of liquid  is determined by the mobility of molecules, which largely depends on the intermolecular forces, and

  1 / 0 ,

(8.4)

where  0 is the molecule mobility, i.e. the of a molecule under the

influence of a single power, and 0   / F , where  is the velocity 98

acquired by the molecule under the action of force F. Einstein proved that the mobility of molecules is proportional to the diffusion coefficient  0  D / kT . Hence, (8.5)   kT D . Let us find the diffusion coefficient D. According to Frenkel’s ideas, the mechanism of diffusion in the liquid can be explained as follows. Denote the mean free path of the particle in fluid by  . In fact,  is the distance between pseudokernels. Then the average velocity of particles within the fluid     /  , where  is the average time of the settled life. Using the well-known formula for the diffusion coefficient of gases: D

1    l , 3

(8.6)

where    is the arithmetic average velocity of molecules,  l  is the mean free path, so we can write for liquid: D

1  2  2 U a / kT .  e 3  3 0

(8.7)

We see that with increasing temperature T the diffusion coefficient increases. This, obviously, can be explained by a sharp reduction in the relaxation time. With regard to the viscosity, using (8.5) and (8.7) we get: U kT a (8.8)   2 e kT or



  ATeU

a

/ kT

.

(8.9)

Thus, with increasing temperature the fluid viscosity decreases rapidly due to the exponential factor. On the other hand, the pre-exponential factor AT depends on the temperature and increases line99

arly with its growth. However, a change in this term is of secondary importance compared with the change in the exponential term, while the values of T are not too high. If the temperature becomes so high that eU a / kT 1 (т.е. kT  U a ), this means that the molecules of the liquid are almost not retained when passing from one pseudonucleus to another, and their movement acquires the same character, as it has in gases. Under such conditions  starts to increase with increasing T, i.e. as well as in gases. When (U a / kT )  1 , i.e. eU a / kT  102103, the factor A is of secondary importance compared with the exponent and the fluid viscosity  decreases with increasing temperature. This relationship (8.9) of the viscosity on temperature is in good agreement with the experimental data within a certain temperature range from the melting point Tпvel to the critical temperature Tkr above which liquid cannot exist. 8.3. Experimental setup for studying temperature dependence of the viscosity of liquids To study the temperature dependence of viscosity of liquids we used the thermostat, in the working tank of which viscometers with tested liquid are mounted. A general view of the thermostat is shown in Fig. 8.1.The setup (Fig.8.1) consists of viscometers 1 and 2 with the tested liquid set by means of special flanges in the plate holes and immersed in the thermostat-regulating medium. The tested liquid in viscometers is blown by air with the help of rubber blowers. The time during which the liquid flows out from the tanks is measured with a stopwatch. As a thermostatic liquid in the thermostat, distilled water, the volume of which is about 16 liters, is used. The water is in the bathtub of the third thermostat. At the bottom of the bath of the thermostat there is an electric heater 4 and a cooling element 8, in which running water, used for cooling of the thermostat medium, circulates. For water mixing, the bathtub of the thermostat is provided with an electromechanical mixer 5, which is to be periodically switched during the experiments. Visual control of the temperature of the thermostatic fluid is carried out by means of a technical thermometer. To do this, contact 100

6 and 7 control mercury thermometers are mounted on the thermostat cover. The thermostat allows us to smoothly change the temperature from room temperature (or running water temperature) to 50 °C.

Fig. 8.1. General view of the thermostat

To measure the viscosity of the fluid in this laboratory work we use standard Ostwald-type Pinkevich VPZh viscometers. IWF viscometer is a U-shaped glass tube (Fig. 8.2.), in one knee of which capillary 6, passing into tanks 4 and 5, is soldered. In the lower part of knee 2 there is an extension 7. When measuring the liquid viscosity, the liquid flows from the measuring container 4 through the capillary 6 to the expansion 7. The rubber bulb 8 provides the fluid movement in the viscometer.

Fig. 8.2. Viscosimeter VPZh

101

According to the Poiseuille law, the distribution of the fluid velocity over the cross section in the laminar flow in a capillary tube of circular cross section far from the entrance to the tube has the following form: p  p2 2 (8.10) R  r 2  ,  1 4 l

where p1  p2 is the pressure drop across the length of the tube section, R is the radius of the tube, r is the distance from the tube axis to the point in question, where the flow rate is equal to  ,  is the viscosity of the liquid flowing through the tube. Knowing the velocity distribution over the cross section of the tube, the volume of fluid Q flowing through the tube cross-section over time  can be determined from the following relationship: R

Q      2  rdr  0

p1  p2  R 4 . 8   l

(8.11)

The principle of operation of the VPZh viscometer is based on the Poiseuille law (formula (8.11)). The difference in liquid levels in the tank 4 and expansion 7 is typically h ≈ 200 - 250 mm. The liquid is filled so that the free liquid level in the knee 2 is within the extension 7. During the measurements, the liquid level in the knee 1 decreases from the top mark M1 to the bottom mark M2 on the measuring reservoir 4. The distance between the marks is h ≈ 15 – 20 mm. As the fluid moves under its own weight, we can write: p1  p2   g h ,

(8.12)

where  is the density of the tested liquid, h is the difference between the liquid levels in the knees of the viscometer. Neglecting the change in h during fluid motion (h  h) , we can rewrite the expression (8.11) for the volume Q taking into account (8.12) as: 102

Q

R 4    g  h   , 8  l

(8.13)

where Q is the volume of the measuring tank 4 between the marks M1 and M2, τ is the time needed for this fluid to flow off the volume. Then, using the expression (8.13), we find the viscosity of the tested liquid:  R4 g h . (8.14)  8Q l The last expression can be written as:   B ,

where B

 R4 g h . 8Q l

(8.15) (8.16)

In this expression all the quantities in the coefficient B are constant for the given design of the viscometer and B is a constant of the device, the value of which is recorded in the passport of each viscometer. Thus, in order to determine the viscosity of the tested liquid, it is sufficient to know its density at a given temperature T, the constant of the device and the time of fluid running out from the measuring reservoir 4 of viscometer VPZH. 8.4. Experimental technique Measurements of fluid viscosity via VPZh viscometer were made as follows. Close the end of knee 2 with the finger, and with the rubber bulb 8, attached to the appendage 3 of the wide knee 2, drive the tested liquid into the narrow knee, to about the middle of the upper reservoir 5. Then remove your finger from the wide knee 2 and watch lowering of the liquid meniscus in the knee 1. When the liquid meniscus reaches the mark M1, switch on the stopwatch, and at the moment when the meniscus passes past the mark M2, switch off the stopwatch. Use the stopwatch to measure the time of the liquid out103

flow  , and use the formula (8.15) to determine the viscosity of the liquid. 8.5. Procedure of work fulfillment 8.5.1. Check the position of toggle switches (mixer, «power» and the heater switch). Toggle switches must be in the lower position («off»). When the setup is switched on, the red lamp «power» lights. 8.5.2. Check the hose connection of the refrigerator with the water supply and the sink. 8.5.3. Fix the temperature of the control thermometer. It shows the room temperature. Make the first measurement of time of fluid outflow from the control volume at room temperature. 8.5.4.Switch on the setup. 8.5.5. Switch on the mixer. Then the electric motor is switched and the circulation of water in the bath starts. 8.5.6. Switch on the heater. To do this, set the switch «heater» in the upper position. The water temperature in the bath will begin to rise. 8.5.7. Upon reaching a reference temperature, turn off the heater and measure the time of liquid flowing out from the control volume. 8.5. The liquid temperature should be raised no higher than 40 °C. Then switch off the heater. 8.5.8. By raising the temperature of water in the thermostat with a heater with a step (12) °С in the range of (2040) °С, determine the time of liquid flowing out from the reference volume (from mark M1 to mark M2) in the viscometer. Measure the time for each temperature at least three times. Write down the measurement results in Table 8.1. Table 8.1 Measurement of the time of liquid flowing out in the viscometer

tbeg ,0 C

t fin ,0 C

 t ,0 C

1, c

2,c

 3, c

  , c

8.5.9. Using the parameters of viscometers given in the data sheet of the device, find viscosity values  of the tested liquids for 104

all temperature values, using the formula (8.15). Write down the results of calculations in Table 8.2. Table 8.2

Calculation of liquid viscosity coefficients  ,10 2 Pа  с

 t ,0 C

 5 10 T

1 ,10 3 K 1 T

T, K

ln

 T

8.5.10. By the results of measurements, plot the temperature dependence of viscosity   f (t ) , where t is the temperature in °С. 8.5.11. Using the experimental data, calculate the values, and plot the dependence ln( / T )  f (1 / T ) , using formula (8.9). To do this, transform the formula. Divide the two parts of the formula by T , find the logarithm and then you will get: U 1   ln    ln A  a  . T k T  

(8.17)

Formula (8.9) is reduced to the linear dependence y  a  bx , where Ua   , x 1. (8.18) y  ln   , a  ln A , b  k T T  It follows that U a / k is determined as an angular coefficient of the slope of the straight line in coordinates ln( / T )  f (1 / T ) . Parameters of (8.17) can be calculated by the least square method using table 8.3. Formulas for determining parameters of the linear dependence by the least squares method: n

a

n

n

n

x x y x  y i 1

i

i 1

i

i

2

i 1

2 i

i 1

n  n    x i   n x i2 i 1  i 1 

n

i

,

b

n

y x i 1

i

2

 n x i y i i 1

    x i   n x i2 i 1  i 1  n

105

i 1

n

i

n

. (8.18)

In table 8.3. and formulas (8.18) i is the ordinal number of the experiment. Table 8.3

Data for calculation of parameters of the dependence U 1 (8.17)   ln    ln A  a  k T T  №

xi 

1 T

  y i  ln   T 

xi2

xi y i

1 2 . n n

x i 1

i



n

y i 1

i

n

x



i 1

2 i



n

x y i 1

i

i



5.8.12. Substituting the value of the Boltzmann constant k, determine the activation energy U a . Compare the resulting value of the activation energy with the electron work function in metals and estimate their order. 8.5.13. Find the value of the coefficient A in the formula (8.9), using the condition (8.18) a  ln A . Then A  e a . 8.5.14. Knowing the value of A, write the final form of the semiempirical formula for the temperature dependence of the viscosity of tested liquids. 8.5.15. Based on your measurements make conclusions, in particular, about the applicability of the Frenkel formula (8.9) for the description of the temperature dependence of the viscosity of the liquid. Questions for self-assessment 1. For what movement of the liquid layers can the Poiseuille formula (8.11) be used? 2. What is called the viscosity and what is its dimension? 3. Why are capillary viscometers used? 4. Why is an expansion between marks М1 and М2 (see Fig.8.2) used in the narrow knee viscometer? 5. Does of the height of liquid level in the expansion 7 affect the outflow velocity? 6. Explain the nature of the thermal motion of molecules in liquids. 7. What is the activation energy?

106

8. What is the time of «resident» life of molecules? 9. What is the nature of changes in the viscosity of liquids and gases when the temperature increases? What is the difference and what is it caused by? 10. Why do we need a thermostat and how does it work? References 1. Matveev A.N. Molecular Physics: a Textbook for physical faculties of universities. – 2 nd ed., Revised and ext. / A.N. Matveev. – M.: Vysshaya shkola, 1987. – P. 360. 2. Kikoin A.K., Kikoin I.K. Molecular physics. – 2nd ed. / A.K. Kikoin, I.K. Kikoin. – M.: Nauka, 1976. – P. 480. 3. Savelyev I.V. The course of general physics. Bk. 3: Molecular Physics and Thermodynamics / I.V. Savelyev. – M.: Astrel. AST, 2003. – P. 208. 4. Frenkel Ya.I. Kinetic theory of liquids / Ya.I. Frankel. – M.: Publishing House of the USSR Academy of Sciences, 1977. – P. 592. 5. Laboratory works in physics: Textbook / L.L. Goldin, F.F. Igoshin, S.M. Kozel et al..; Ed. Goldin L.L. – M.: Nauka. Main publishing house of Physical and Mathematical Literature, 1983. – P. 704. 6. Physical practicum. Mechanics and Molecular Physics: textbook / Iveronova V.I., Belyankin A.G., Motulevich G.P. Chetverikova E.S., Yakovlev I.A. Ed. Iveronova V.I. – M.: Nauka, 1967. – P. 352. 7. Zeidel A.N. Errors of measurements of physical quantities / A.N. Zeidel. – SPb.: Lan, 2005. – P. 106.

107

Laboratory work →9 DETERMINATION OF THE SURFACE TENSION COEFFICIENT OF LIQUIDS 9.1. The purpose of work: get acquainted with the method of experimental determination of the surface tension.

9.2. A brief theoretical introduction Molecules of the liquid are so close to each other that forces of attraction between them have rather high values. The force of interaction decreases rapidly with distance. In the first approximation, we can assume that the force of attraction is F ~ 1 , and the force of rer6

pulsion is F ~ 112 , where r is the distance between molecules. r

Fig. 9.1

Starting with a certain distance, the forces of intermolecular interaction can be neglected. This distance r0 is called the radius of molecular action and a sphere of radius r0 is called a sphere of molecular action. Each molecule is attracted by all the neighboring molecules within the sphere of molecular action, the center of which coincides with this molecule. The resultant of these forces for the 108

molecule inside the liquid at a distance r0 from the liquid surface is on average equal to zero (Figure 9.1). The situation is different if the molecule is at a distance from the surface less than r0 . Since the density of the gas is much less than the liquid density, the part of the sphere of molecular action outside the liquid will be less filled with molecules than the other part of the sphere. As a result, force directed into the liquid will act on each molecule in the surface layer of thickness r0 ,. The magnitude of this force is increasing from the inner to the outer boundary layer. Therefore, when the molecule moves from the liquid depth to the surface layer, it must do work against the forces acting in the surface layer. This work is done by the molecule at the expense of its kinetic energy is used to increase the potential energy of the molecule. When the reverse transition of the molecule occurs, its potential energy is converted into kinetic energy. Thus, the molecules in the surface layer have more potential energy. The value of the potential energy, or more precisely, the free energy E of the surface layer depends on the number of the surface molecules, i.e., the surface area S . This energy is called the surface tension energy. It can be written as:

E  S ,

(9.1)

where  is the coefficient of proportionality, called surface tension. The surface tension coefficient  is numerically equal to the free energy per unit area of the surface layer. Note that the value of  depends on the types of two boundary media. We can give another definition of the surface tension coefficient. The surface tension coefficient  is numerically equal to the work to be done to increase the area of the surface layer by unit. The forces of surface tension strongly affect the behavior of liquids. In particular, the shape taken by the liquid is determined by the minimum potential energy, which is the sum of the energy of surface tension and the potential energy in the gravity field. 109

In calculations, instead of the surface tension energy we often use the surface tension force, i.e., the force with which one part of the liquid surface acts on the other part. Along any line «aa», mentally drawn on the surface of the liquid (Figure 9.2) act forces perpendicular to the line and tangent to the surface of the liquid, these forces «pull together» sections I and II separated by the line.

Fig. 9.2

These forces are the greater, the longer the line «aa» is: F  l,

(9.2)

where l is the length of line «aa». The free energy increases when the surface area increases because the work against the forces of surface tension is done. According to formula (9.2), the surface tension coefficient is equal to the force of surface tension acting tangentially to the surface of the liquid per unit length of the section line. The surface tension coefficient is measured in SI in N / m or J//m2. Many physical phenomena in liquids, such as waves on the liquid surface, wetting and non-wetting phenomena, capillary phenomena are explained by surface tension. Experiments show that if the end of the capillary is immersed in the liquid poured in a wide vessel, the pressure under the curved surface in the capillary will differ from the pressure in the vessel flat surface vessel by an amount p , determined by Laplace's formula:  1 1 p      R1 R2 110

  , 

(9.3)

where R1 and R2 are radii of curvature of the liquid surface in two mutually perpendicular cross sections. The half-sum

1 1 1    2  R1 R2

   

is called the mean curvature of the surface in that point. For all forms of surfaces, which can form in the liquid, the mean curvature remains constant for any pair of mutually perpendicular cross-sections normal to the surface in the given point. Therefore, these sections are chosen for reasons of convenience. For a spherical surface R1  R2  R , where R is the radius of the sphere, therefore, for the round capillary the formula (9.4) takes the form: p 

2 . R

(9.4)

As a result, in case of wetting of the capillary walls, the liquid level therein is higher than in the vessel, in case of nonwetting – it is lower. Changing of the height of the liquid level in narrow tubes is called capillarity. Such a level difference h is established between the liquid in the round capillary and the wide vessel that the hydrostatic pressure  g h is balanced the capillary pressure p : 2 R  gh ,

(9.5)

where  is the liquid density;  is the surface tension coefficient on the liquid – gas surface. The radius of curvature of the meniscus R can be expressed in terms of the radius of the capillary r , and the contact angle  . As seen from Fig.9.3, r  R  cos . Then h

2  cos  .   g r

(9.6)

In case of full wetting   0 , cos   1. Then the radius of curvature R is equal to the radius of capillary r and the formula (9.4) takes the form: 2 . (9.7) p  r

111

Fig. 9.3. To the derivation of formula (9.6)

For the height of liquid rise h in the capillary we have: h

2 .   g r

(9.8)

We can find the coefficient of surface tension  :



  g hr . 2

(9.9)

Hence, it is sufficient to measure experimentally the liquid column height h in the capillary and capillary radius r, and knowing liquid density  , calculate  according to the formula (9.9). 9.2 The experimental setup and measurement technique To measure the height of the liquid column in the capillary it is necessary to use high-precision instruments. Therefore, in this work, this measurement is replaced by the measurement of the excess of the capillary pressure p . The method of compensation of additional pressure is used. To measure the pressure, a multirange micromanometer with an inclined tube of MMH type is used. The description and the operating principle of the multirange micromanometer with an inclined tube is given in the appendix at the end of the manual. If we put a capillary tube in the vessel with the tested liquid, in case of wetting of the walls of the tube with the liquid, the latter rises 112

in the capillary to a certain height due to the additional pressure (in this case negative). The additional capillary pressure in the round capillary in the case of complete wetting is defined by (9.7). If by one or another way we increase the external pressure on the surface of the liquid in the capillary, it is possible to reach such a state when the liquid level in the capillary and in the wide vessel are equal. It is obvious that in this case the excessive external pressure pint is equal to the additional pressure p , determined by the formula (9.7). Therefore, compensating the additional pressure, we can measure it and determine the surface tension. If the external pressure is measured in pascals by micromanometer, then (9.10) pint  9,804 kPa , where n is the length of the alcohol column in the measuring tube of the micromanometer in divisions; k is the coefficient of inclination of the measuring tube. From (9.7) and (9.10) we get:   9,804

k nr . 2

(9.11)

The setup (Fig. 9.4) has a capillary 1, which through the threeway valve 2 is connected to the micromanometer 4 through the tube 3 and to compensator 6 through pipe 5. To increase the ambient pressure above the liquid surface in the capillary, i.e. to compensate for the additional pressure compensator 6 is used. The compensator is a cylinder with a rubber bulb placed inside it. Rotating the compensator clockwise, we squeeze the bulb, thereby increasing the external pressure on the surface of the liquid in the capillary. Through the tube the three-way valve tube 2 is connected to the rubber bulb 7, which is used to remove residual tested liquid from the capillary. In this work the surface tension coefficient of different concentrations of alcohol solutions is determined. 113

1 – capillary, 2 – three-way valve, 3 – tube, 4 – micromanometer, 5 – tube, 6 – diaphragm pump, 7 – rubber bulb, 8 – tripod 9 – post Fig. 9.4. Setup for determining the surface tension coefficient of liquid

9.3. Procedure of work fulfillment 9.3.1. Pour distilled water in a glass, put the capillary in it and rinse it with this liquid. 9.3.2. Set the cup so that the end of the capillary was immersed in the liquid. The liquid rises in the capillary to a certain height (the column of liquid should not contain air bubbles). 9.3.3. Slowly rotate the handle of compensator 6 clockwise, raising the pressure in the system until the fluid level in the capillary lowers to the liquid level in the cup. 9.3.4. Record readings of micromanometer. Repeat measurements 3 ÷ 5 times. Find the average n value. Wrote the data in table 9.1. Table 9.1 Measurement of excess pressure in alcohol solutions of different concentrations с, %

k

n1 , div

n2 , div

114

n3 , div

 n , div

9.3.5. Make similar measurements with aqueous solutions of alcohol С2Н5ОН, starting with a low concentration of alcohol, then take solutions with higher concentrations. Write the in Table 9.1. 9.3.6. Calculate the additional pressure p and surface tension coefficients  for aqueous alcohol solutions of different concentrations using formulas (9.10) and (9.11). Write the calculation results in Table 9.2. Table 9.2 Calculation of additional pressure p and surface tension coefficient  с, %

 n , div

,N /m

p, Pa

9.3.7. For measurements with water, calculate the random error  ñë by the method of finding errors of direct measurements, using the data in Table 9.3. Table 9.3 Calculation of random error  ran for measurements with water №

ni , div

p i ,

i,

Ïà

Í /ì

  , Í /ì

1 2 3

 i ,

 i2 ,

Í /ì

Í

2



2

9.3.8. For one of the dimensions estimate the instrument error

 error using the method of finding errors for indirect measure-

ments:

2

2

 r   n   error        ,  r   n 

(9.12)

where r is the error in the determination of the capillary radius (indicated); n is the division value of micromanometer scale, 1 mm. 9.3.9. Estimate the error of the result taking into account random and instrument error using the formula: 115

2 2 .  full   ran   error

(9.13)

9.3.10. Estimate the relative error:

   

 full  

.

Present the final result as:       full ,

Н. м

9.3.11. Plot the dependence of the surface tension of the liquid on the concentration of alcohol in it. 9.3.12. Determine the unknown concentration of the solution using the graph. Questions for self-assessment 1. What causes surface forces in liquids? 2. What are the phenomena of total wetting and total non-wetting? 3. What is called the coefficient of surface tension? What does it depend on? 4. What causes additional pressure? 5. Write Laplace formula for additional pressure and explain it. 6. Explain capillary phenomena in liquids. 7. What tubes can be considered capillary? What is the height of liquid rising in the capillary determined by? 8. Tell about the method of pressure compensation. What pressures balance each other and what are they are caused by? References 1. Physical practicum. Mechanics and Molecular Physics. Ed. Iveronova V.I. – M.: Nauka, 1967. – P. 352. 2. Matveev A.N. Molecular Physics: a textbook for physical specialties of universities. – Ed. 2nd, Revised and compl. – M.: Vysshaya shkola, 1987. – P. 360. 3. Kikoin I.K., Kikoin A.K., Molecular Physics. – M.: Nauka, 1976. – 480 p. 4. Zeidel A.N. Errors of measurement of physical quantities / A.N. Zeidel. – SPb.: Lan, 2005. – P.106.

116

Laboratory work →10 DETERMINATION OF THERMAL CONDUCTIVITY OF SUBSTANCES BY COLORIMETRIC METHOD IN THE QUASI-STATIONARY REGIME 10.1. The purpose of work: 10.1.1. Become familiar with the phenomenon of energy transfer in the substance. 10.1.2. Master the colorimetric method of measuring thermal conductivity of substances.

10.2. A brief theoretical introduction Thermal conductivity A region of space filled with some substance, each point of which has a value of temperature T  T x, y, z, t  depending on coordinates and time, is a temperature field. It is non-stationary if temperature depends on time and stationary if temperature depends only on coordinates T  T x, y, z  . If the temperature is the same at all points of the field, the temperature field is uniform. In the nonuniform temperature field thermal energy is always transferred from areas with higher temperature to areas with lower temperature. There are three mechanisms of heat transfer. 1. Convection is heat transfer by a moving medium. Convection is possible in liquid and gaseous media. 2. Radiation is heat transfer by electromagnetic waves. 3. Thermal conductivity is a direct transfer of kinetic energy of molecular motion of microparticles to adjacent microparticles of the same body or another body, which is in thermal contact with the first. In this work the process of thermal conductivity is studied. Heat flow If there is a temperature difference between two neighboring points of the temperature field, the temperature will be equalized by 117

conduction. In a conventional sense, «heat» «flows» from one area to another. Therefore, the amount of heat passing from one part of the body to another section through a unit area per unit time is called the heat flux density q. Fourier (Fourier Jean Baptiste Joseph, 1768 - 1830) empirically established that the heat flux density is equal to:

 q    grad T , where  is thermal conductivity of the substance; grad T , a temperature gradient is a vector directed towards the  greatest increase in temperature, i.e. along the normal n to the isothermal surface and numerically equal to the derivative of the temperature in this direction:  T  T  T  ; k j i grad T  T  x y dz

grad T  T / n .

In case of one-dimensional temperature field T  T x  , the gradient turns into the derivative of temperature with respect to x, i.e. q  

dT . dx

(10.1)

From (10.1) it is easy to see that the thermal conductivity  is numerically equal to the amount of heat passing through a unit area perpendicular to the direction of heat flow per unit time when the temperature gradient is equal to unity. In SI  is measured in W/(m∙K). For small temperature intervals  is independent of temperature. In the general case  is a function of temperature, pressure, etc. If the heat transfer process is stationary and the temperature varies from layer to layer uniformly, the ratio (10.1) is written as: q

T1  T2 , x 118

(10.2)

where T1  T2 is the temperature difference between the two cross sections of the substance; x is the distance between them. In the description of thermal conductivity, the coefficient of thermal conductivity is used.  , a c where c is the specific heat of the substance,  is its density. Thermal conductivity, coefficient of thermal conductivity and specific heat of a substance are of great theoretical and practical significance. Ii is very important to know these characteristics in all branches of engineering using thermal processes. There are many methods for determining these characteristics. We will study one of them in more detail. Let us consider a one-dimensional problem of heat transfer through a layer of thickness x. Suppose that on one side of the layer constant temperature TH is maintained, and on the other side there is a second medium with the temperature T  TH , not having a heat exchange with the external environment. The heat Q transmitted through the second partition will cause a change in the temperature of the second medium: dT 

Q , cm

(10.3)

where c is the specific heat capacity of the second medium; m is the mass of this medium. The amount of heat Q passing through the area S during time d is: Q  q  S  d . If we can assume that the temperature in the layer changes by a linear law (the so-called quasi-stationary mode), then

Q  

T  T  S  d . x 119

(10.4)

Then, based on (10.3) and (10.4), we can write: cm x

dT    S  d . TH  T

(10.5)

If the medium temperature behind the wall during some time in this mode has changed from T1 to T2 , the relation (10.5) can be written as: 

T2

dT    S   d . T  T T1  0

c  m x

(10.6)

After all the mathematical operations, we will get the expression for the thermal conductivity of the partition material in the form:



c  m  x T  T1 . ln S  T  T2

(10.7)

Thus, if we know the mass m and specific heat c of the medium of the partition, its initial temperature T1 and final temperature T2 , the time  during which this temperature change in the medium occurred, thickness of the wall x, the area of conducting surface S and temperature TH , then it is not difficult to determine the value of thermal conductivity of the substance from which this partition is made. 10.3. Description of the experimental setup A schematic diagram of the experimental setup is shown in Figure 10. Steam is continuously supplied from the tank 6 to the reservoir for steam 7, which provides conditions of constant temperature TH . A layer of bulk material (sand) 2, the thermal conductivity of which is to be determined, is placed on the metal plate covering the reservoir for steam. Calorimeter 1 with water is put on the sand layer. The reservoir for steam, the layer of bulk material and the calorimeter are heat-insulated from the environment. 120

1 – calorimeter 2 – bulk material, 3 – insulating sleeve, 4 – cover, 5 – thermometer, 6 – water tank, 7 – reservoir for steam, 8 – tube, 9 – electric heater, 10 – plug, 11 – power cord, 12 – tank neck, 13 – stopper, 14 – level indicator, 15 – drain hole, 16 – screw, 17 – mixer. Fig. 10. Setup for determining thermal conductivity of bulk materials

Changes in the water temperature in the calorimeter are registered using thermometer 5. If we know thickness of the bulk material layer x, its area S, steam temperature TH and water temperature in the calorimeter at the initial moment of time and after a period of time  , we can calculate the thermal conductivity from the relation (10.7), which for our setup cam be written as:



c1m1  c2 m2 T  T1 ;  x  ln  S  T  T2

(10.8)

here c1 and c 2 are specific heat capacities of water and vessel;

m1 and m 2 are their masses. 10.4. Procedure of work fulfillment 10.4.1. Weigh the empty calorimeter and calorimeter with water; find the mass of water in the calorimeter. 121

10.4.2. Measure the thickness and area of the layer of bulk material. Write the results of measurements in pp.10.4.1-10.4.2 in Table 10.1. Parameters of the experimental setup Calorimeter

Mass, kg Calorimeter with water

Water

Table 10.1

Size of studied samples, m Thickness  x  Diameter d  x d

10.4.3. Assemble the experimental setup in accordance with the scheme. 10.4.4. Determine the boiling point of water in the boiler (from the table for the given atmospheric pressure) and measure the initial water temperature in the calorimeter (it is desirable that it is 5-10 degrees below the ambient temperature). 10.4.5. Turn on the heater and wait for steam going into the vessel 7. Start measuring time and the water temperature in the calorimeter. 10.4.6. Measure the dependence of water temperature in the calorimeter on time. Make measurements until the water temperature in the calorimeter is higher 5-10 degrees higher than the room temperature. Write the data in Table 10.2. Table 10.2 Dependence of water temperature in the calorimeter on time

 , мин t, 0C

10.4.7. Using the data of measurements plot the graph T  f   . The temperature values T and the corresponding time intervals  used in calculations in formula (10.8) are determined from the linear part of the graph. 10.4.8. Using the results of the experiment determine the thermal conductivity of samples.

122

Questions for self-assessment 1. What is thermal conductivity? 2. What is heat flux density? 3. How are the heat flux density and the ambient temperature interconnected? 4. What is the physical meaning of thermal conductivity of the substance? 5. What does the term «quasi-stationary mode» mean? 6. Why is the method of determining thermal properties of a substance used in this work called calorimetric method? 7. Measurement of which quantity contributes most to the error of the result? References 1. Matveev A.N. Molecular Physics: a Textbook for physical faculties of universities. – 2 nd ed., Revised and ext. / A.N. Matveev. – M.: Vysshaya shkola, 1987. – P. 360. 2. Kikoin A.K., Kikoin I.K. Molecular physics. – 2nd ed. / A.K. Kikoin, I.K. Kikoin. – M.: Nauka, 1976. – P. 480. 3. Savelyev I.V. The course of general physics. Bk. 3: Molecular Physics and Thermodynamics / I.V. Savelyev. – M.: Astrel. AST, 2003. – P. 208.

123

APPENDIX APPENDIX 1 Derivation of the binomial distribution Let the experiment be made with four coins: A, B, C, D. Let us suppose that in one of the experiments, all the coins fell «eagle» up – the state of «E» (k = 4). Then this arrangement can be represented as a sequence of A (E) B (E) C (E) D (E). According to the rule of multiplication of probabilities of independent events, the probability of this state is pppp  p 4 . Such a state can be realized in only one way, as the coins in this state may be arranged in any order. Consider the case when in the other experiment any three coins were in the state «E» (Eagle) and one – in the state 'T' (tails). There will be four variants: A (E) B (E) C (E) D (T), A (E) B (E) C (T) D (E), A (E) B (T) C (E) D (E), A (T) B (E) C (E) D (E). Since the probability of each of these states is ppp  p 3 , the probability of one of the listed options, according to the theorem of addition of probabilities of independent events, is equal to 4 p 3 . For (k  2) the number of possible options of arrangement in two states «E» and «T» will be equal to six, and the probability of the state is equal to 6 p 2 q 2 (show it). The results are summarized in Table A1. Distribution of coins for two states (n = 4, p = 0.5) No. of states 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 

State «E» ABCD ABC ABD ACD BCD AB AC AD CD BD BC D C B A -

«T» D C B A CD BD BC AB AC AD ABC ABD ACD BCD ABCD

Table A1

k

C 4k

p k q 4 k

wk 4; 0,5

4 3

1 4

16 16

1/16 4/16

2

6

16

6/16

1

4

16

4/16

0

1 16

16

1/16 1

124

It is easy to notice that the number of ways of placing coins in this state («E») is equal to the numerical coefficient in the binomial expansion (binomial theorem): ( p  q) 4  1 p 4  4 p 3 q  6 p 2 q 2  4 pq 3  1q 4 .

It is known that the binomial coefficients of this expansion are equal to the number of combinations of 4 elements by k = 4,3,2,1,0 (check the data in Table A1):

C4k  4! k!(4  k )! Each of the components of the binomial determines the probability of a particular state, i.e.: wk 4 , p   C4k p k q 4k . If this formula is valid for n  4 , it can be applied to n  5,6,..., etc. (method of mathematical induction). Then for n coins this expression takes the following form: wk n , p   Cnk p k q n k . This relation is called the binomial distribution. It determines the probability that in n tests the event we are interested in (falling «eagle» up) will happen k times. With regard to the physical system, which can be in two states, the binomial distribution allows us to calculate the probability of a given state.

125

APPENDIX 2 Characteristics of the binomial distribution 1. Mathematical expectation Let us find out mathematical expectation for the binomial distribution. Substituting (1.1) into the formula (1.2), we obtain: n

mk   k k 0



n! n! p k q n k  0  pq n1  k ! n  k  ! 1! n  1 !

n! n! p n q0 . p 2 q n  2  ...  2! n  2  ! n!0!

This expression can be simplified using the properties of the factorial: n! n(n  1)! , n! n(n  1) (n  2)!,…

Then we obtain:

mk  npq n1  n( n  1) p 2 q n 2 ... p n    np qn 1(n 1) pqn 2 ... p n 1  np(q  p)n 1np.  

 

It was taken into account that (q  p) n1  1 as q  p  1 . Therefore, the sought expression for the mathematical expectation of the binomial distribution is written as: mk  n p . 2. Variance Let us determine variance of the binomial distribution, using its mathematical definition (1.4):  k  n k 0 n n k  k    k 2  2kmk  m k2 W  ; p   k 2W  ; p   n  k 0 n  k 0 n n k  k  2  2mk  kW  ; p   mk  W  ; p . n  n  k 0 k 0 n

 k2   k  mk 2 w ; p ,

 k2

The first term in this expression is the mathematical expectation for k 2 , the sum in the second term is the mathematical expectation for k, and the sum in the third term is the probability that out of n tests the «Eagle» will fall out 0, 1, 2, ..., n times. Obviously, this probability should be equal to 1, as falling out of any number of «eagles» in n tests is a certain event: n

 wk

n , p   1.

k 0

126

This expression is called the normalization condition. Taking this into account, the expression for the variance can be written as follows:

 k2  mk  2mk2  mk2  mk  mk2  mk  (np) 2 . 2

2

2

To calculate mk we will use the equality:

k 2  k 2  k  k  k (k  1)  k . Then

mk 2  mk ( k 1)  mk  mk ( k 1)  np.

Substituting this relation in the expression for the variance, we get:

 k2  mk ( k 1)  np  (np) 2 . Using the definition of mathematical expectation for function k (k  1) , we obtain: n n! mk ( k 1)   k ( k  1) p k q n k  k !(n  k )! k 0 n! p 2 q n2 n! p 3q n3 n! p n  3 2 L n(n  1)  2 !(n  2)! 3!(n  3)! n ! 0!  n (n  1) p 2 q n2  n (n  1) (n  2) p 3 q n3    n (n  1) p n 

 002





 n(n  1) p 2 q n2  (n  2) pq n2    p n2  n(n  1) p 2 , as the sum in square brackets is equal to one: q n 2  (n  2) pq n 2 L  p n 2  (q  p) n 2  1.

Therefore, the expression for the variance of the binomial distribution can be written as:  k2  n(n  1) p 2  np  (np)2  (np)2  np 2  np  (np)2  np(1  p.)

Taking into account that 1  p  q , we get:

 k2  n p q.

127

APPENDIX 3 Derivation of the Laplace-Gauss distribution Transform the binomial distribution using Stirling's formula for the factorial: n!  2nn n e  n , which is valid for n  10 . Then



n! k  w ; p   p k q n k   n  k!(n  k )! 2nn n e  n

2k k k e  k

 

1 2

2 (n  k ) (n  k ) n  k e  ( n  k )

p k q n k 

n n n k  k p k q n k  k k (n  k ) k (n  k ) n  k

1 2

k

n  np   nq      k (n  k )  k   n  k 

n k

.

Let us introduce the variable u  k  n p , and express k through u:

k  u  np, then 1 k  w ; p   n  2

 np  n   (u  np)(n  u  np)  u  np 

u  np

 nq     n  u  np 

n  u  np

.

Taking into account that p  1  q and taking out the brackets np and nq and making further obvious mathematical operations, we get: 1 k  w ; p   2 n p q n 

 u  1   np   u  u  1  1    np  nq  1

 ( np u )

 u  1   nq  

 ( nqu )

.

If n   , thenо u / np  1 and u / nq  1 . Therefore, in the expression under the radical we can leave only a unit. This cannot be done with respect to the last two factors, since they are raised to higher degrees. Let us take logarithms of the last expression:   u  u  k    (n q  u ) ln 1  . ln w ; p    ln 2 n p q  (np  u ) ln 1  n p n q  n    

Expand the logarithmic function into a series by the formula:

128

ln(1  x)  x 

x2 x3  L 2 3

Confining by the first two terms, we obtain: 2

2

 u u 1  u  , ln 1  u    u  1  u  .   ln 1       2 nq  nq    nq   np np 2  np

Substituting these expressions in the initial formula and performing algebraic manipulation, taking into account that p  q  1 , we obtain: ln

wk n ; p  2 n p q



u2 . 2n p q

Hence, going back to the old variable and potentiating, we obtain the desired expression: wk n ; p   1

2 n p q  e



( k n p )2 2n p q

.

This is the Laplace-Gauss distribution. Since in the derivation of this relation the approximate Stirling formula for factorials was used, which is valid only for n  10 , the Laplace-Gauss distribution is also valid only for n  10 . For n   the Laplace-Gauss distribution coincides with the binomial distribution.

129

APPENDIX 4 Derivation of Maxwell distribution The thermodynamic equilibrium is established as a result of the great number of collisions between molecules. As a result of each collision the speed projection of the molecule experiences random changes  x ,  y ,  z , and changes in each projection of the speed are independent of each other. Consider the motion of molecules whose velocity at the initial moment of time is zero. Denote changes in projections of its velocity after the i-th collision with other molecules as  x i ,  ,  z i . After a sufficiently long period of time, yi

during which the molecule experienced an enormous number of collisions, its velocity projections will be equal to:

 x    x i ,  y    y i ,  z  i

i

i  z i .

Each of these projections is the sum of a large number of random variables for which the Gauss law is valid (see., for example, [1], §2). Hence, the probability density function for the velocity components can be represented as follows:

 ( x2 )  Ae

 2x

,  ( y2 )  Ae

 2y

,  ( z2 )  Ae

 2z

,

(П1)

where A and  are constants, the same for all three projections due to compete equivalence of the coordinate axes and independence of random variables  x ,  y ,

z .

The probability that the projection of the velocity on axis x is in the interval

 x , x  d x  is equal to:

 

2

dP x     x2 d x  Ae  x d x .

(П2)

Similar formulas are valid for the other projection of the velocity. The probability that the velocity of the molecule is in the interval

 , x

y , z , x



 d x , y  d y , z  d z ,

by the law of multiplication of probabilities is expressed as:





dP  x , y , z  A 3e

   x2  y2  z2 

d x d y d z .

(A3)

The constant A is determined from the normalization condition:   

   dP ,  ,    1 . x

y



130

z

(A4)

This condition implies the following: The probability that the molecule can have any speed within the interval (, ) is equal to one, i.e., the event is absolutely certain. Taking into account that



e

 x2

d x 



dition (A4) we find:

 , from the normalization con

a.

A

(A5)

Let us calculate the mean kinetic energy of the molecule: m 2 m m   x2   y2   z2      x2   y2   z2 dP x ,  y ,  z   2 2 2  

m     2  

3

2   

   

2 x

  



  y2   z2 exp    x2   y2   z2 d x d y d z .

   

The integrals of this type are solved as follows: 

 e

2 x2 x

d x 



 



e

x2

d x  



  1  23 .  a   a 2

Applying this technique to each integral, we obtain: m 2 3m .  2 4a

Using the relationship between the average kinetic energy of molecules and temperature: m 2 2  3 2 kT , we get expression for :   m 2kT .

Taking into account the resulting expression and formula (A5), the equation for the probability (P3) can be written as:





dP  x , y , z  m 2kT 

32

e



m  x2  y2  z2  2 kT

d x d y d z .

(A6)

From the distribution of velocity projections let us move to the distribution of the velocity module. As the distribution of velocities is isotropic, we must use the system of spherical coordinates in the velocity space and integrate the equation (A6) over the spherical layer of d thickness, the radius of which is equal to   ( x2   y2   z2 )1 / 2 . We obtain:

131

d x d y d z   2 d d ,

where d is the solid angle at which the element spherical surface layer is seen from the origin of coordinates. Obviously, the integral over the entire spherical surface of the layer is equal to:

  d    d  4 2

  4

2

2

.

4

Therefore, as a result of integration in (A6) over the spherical layer of d thickness we obtain the probability of finding the module of the molecule speed in the speed interval  ,   d  :

dP   4 m 2kT  e



m 2 2 kT

f    4 m 2kT  e



m 2 2 kT

32

 2 d .

The function 32

2

(П7)

is called density of Maxwell probability distribution. As the molecules move randomly and independently, the number of molecules dn( ) , whose velocities lie in the interval  ,   d  , is: dn( )  n  dP( ) , where n is the total number of molecules in the system. The relative number of molecules in the interval  ,   d  is: dn   dP   f    d , n

or with account for (A7), we get the Maxwell distribution of molecular velocities: m 2

 dn( ) 3 2  4 m 2 kT  e 2 k T  2 d . n

132

APPENDIX 5 Coefficients A and B in the approximating linear equation (the equation, approximately describing the experimental data) (П.1)

y  A  Bx

calculated by the least squares method using the following formulas: n

n

n

n

 xi  xi yi   xi2  yi ,

i 1 i 1 A  i 1 i 1 2 n  n    xi   n xi2   i 1  i 1 

n

n

 y i  xi

n

 n  xi y i .

i 1 B  i 1 i 1 2 n n     xi   n xi2   i 1  i 1 

(П 2)

If the dependence (2.4) E  at is used, the formula (A.1) takes the form:

y  Bx , where y  E , B  a , x  t . Coefficient B is calculated by the formula: n

 Ei t i .

B  i 1

(A 3)

n

 ti 2

i 1

The values of Ei, ti are taken from the experiment. The formula (2.5) can be written as: E  a  bt , t i.e. reduced to a linear dependence of the form y  A  Bx , where y  E , A  a , t B b, x t . Then, to calculate coefficients A and B we can use formulas (A 2).

133

APPENDIX 6 Thermal emf of thermocouples from different metals and alloys Temperature of «cold» junction is 0 °C. Temperature of «hot» junction, °C

100 200 300 400 500 600 700 800 1000 1500

Table

Thermal emf, mV Platinum, 90% platinum 10% rhodium 0.64 1.44 2.31 3.25 4.22 5.22 6.26 7.33 9.57 15.56

134

Iron, constantan (60% Cu + 40% Ni) 5,4 11,0 16,6 22,1 27,6 32,3 39,3 45,7 58,3 -

Copper, constantan 4,28 9,29 14,86 21,87 -

APPENDIX 7 Molar mass M, density  and pressure of saturated vapors pí at 20 °C for some liquids №

Liquid

1 2 3 4

Water Ethanol Acetone Ethyl ether

Chemical formula Н2 0 С2 Н5ОН (СН3)2СО (С2Н5)2О

5

Isobutyl alcohol

С4Н9ОН

6 7 8 9

Butyl alcohol Benzene Dichloroethane Carbon tetrachloride Ethylbenzene

10

M,

kg  10  3 mol

18,02 46,07 58,08 74,12

,

kg  103 mol

p н , mm рт. ст

0,9982 0,7893 0,7908 0,7135

17,6 42,4 156 440

74,12

0,8027

9

С4Н9ОН С6Н6 1,2-С2Н4Сℓ2

74,12 78,11 98,96

0,8098 0,8790 1,257

5 76 66

ССℓ4 С6Н5С2Н3

153,8 106,2

1,592 0,8670

87 7,1

135

APPENDIX 8 LIQUID DEVICES WITH VISIBLE LEVEL FOR PRESSURE MEASUREMENTS U-shaped double-tube and cup single-tube manometers refer to this type of devices. They are used to measure pressure, vacuum and pressure difference between air and non-corrosive gases.

1 – glass tube, 2 – scale, 3 – h1 , h2 – the height of columns of the working fluid in both legs, measured from the zero of the scale, 4 – h  h1  h2 – the difference between the liquid levels in the legs of the manometer Fig. 1. Scheme of the U-shaped manometer A double-tube manometer consists of a glass tube filled to half of its height with the working fluid, and the scale, allowing us to measure levels in both legs. In case of equal pressures above the liquid surface in both legs, the liquid levels in them coincide with the zero of the scale. Water, mercury and alcohol (with a small inner diameter of the tube) are used as the working fluid. Pressure, rarefication or pressure difference are balanced and measured by the column h  h1  h2 of the working fluid. Manometer is set vertically. To measure the excess pressure in the object, the right leg is connected with the object and the left leg is open (to the atmosphere, p2  patm ): p1  p 2 . When the levels in the legs of the manometer get balanced, the pressures are equal: p2   g h  p1   c g h , Hence

p1  p2  p1  patm  g h   c  ,

where  is the density of working liquid (kg/m3),

 c is the density of the medium over the working liquid (kg/m3), 136

(A1)

g  is the acceleration due to gravity at the latitude of the experiment (m/s2), h is the level difference in the legs of the manometer (m). If  c   , formula (A1) will be written as:

p1  patm   g h ,

(A2)

and the pressure in the object is

p1  patm   g h . When measuring the rarefied medium, the left leg is connected with the object, and the right is open (to the atmosphere), then p1  patm  p2 . For  c   the pressure in the object is calculated by the formula:

p2  patm   g h . In case of measuring the difference in pressure between the two objects, the higher pressure is connected to the right leg and the lower to the left leg of the device, the result of measurement is calculated using the formula (A1). The working fluid density and the gravitational acceleration are given in Tables with error of about 0.005%, hence, the pressure measurement accuracy mainly depends on the accuracy of measurement of the working height of the liquid column h. In the case h is measured with the naked eye, with the scale division of 1 mm and recordings in two legs, the instrument error does not exceed 1mm of the working fluid column. Cistern-type one-tube manometer The manometer consists of a cylindrical vessel 1 and measuring glass tube 3 connected to it. The working fluid is poured into the vessel so that the level in the measuring tube is against the zero scale mark.

1 – glass tube, 2 – scale, 3 – measuring tube FIg. 2. Scheme of cistern-type one-tube manometer

137

When measuring the pressure in the object (the pressure in the object is above atmospheric), it is connected to the wide manometer vessel, and when rarefication is measured in the object (the pressure in the object is below atmospheric) it is connected to the measuring instrument tube. When measuring the pressure difference in two objects, the higher pressure is connected to the vessel, and the smaller – to the measuring tube. If the liquid in the measuring tube rises to the height h1 , and in the wide vessel goes down to a height of about h2 with respect to zero of the manometer scale, then the height of the working liquid column (П3) h  h1  h2 corresponds to the measured value of the pressure difference in mm of the working fluid column. The volume of fluid displaced from the wide vessel is equal to the volume of liquid in the measuring tube, that is

h1 S1  h2 S 2 ,

(A4)

where S1 and S 2 are areas of the inner section of the measuring tube and wide vessel, respectively. Solving Equations 3 and 4, we get:

 S  h  h1 1  1  S 2   and if  c   , according to (A1) we have:

 S  p1  p 2   g  h   g  h1 1  1  .  S2 

(A5)

Cistern-type manometer enables us to make only one measurement of h. For the scale division of 1 mm the recording h1 has an error not exceeding 1 mm of the working fluid column. In measurements the correction for lowering of the liquid level in the wide vessel is typically 0.25% of h1, for

S1 1  this correction is: S2 400

  0,0025h1 , where Δ the error in determination of the working fluid column h1 in mm. Errors in determination  and g  are much smaller and they are neglected. When measuring low pressures, the vacuum or pressure difference more accurate instruments – micromanometers with an inclined measuring tube are used.

138

Micromanometer with an inclined measuring tube Inclination of the measuring tube is made to reduce measurement errors and to increase the accuracy of measured pressure. When measuring the pressure in the object, ( pobject  patm ) it is connected via a hose to the vessel, when rarefied gases are measured ( pobject  patm ) , the measuring tube is connected. In case of measuring pressure difference the higher pressure is supplied to the vessel, and the lower – to the measuring tube. The working liquid is ethyl alcohol with a density   809,5 кг / м 3 , which is poured into a vessel so that the level of it in an inclined measuring tube is against the zero mark of the scale. The scale length is 250  300 mm. Let under the action of the measured pressure the liquid level in the measuring tube, inclined at an angle  to the horizontal plane, (Figure 3) rise vertically to a height h1, and lowers to h2 in the wide vessel. Then the difference of the heights h of working fluid levels in the device, balancing the measured value p, will be equal to: h = h1+ h2.

(П6)

If the liquid column in the measuring tube, measured on the scale is equal to n mm, then (П7) h1  n sin  . Then the volume of liquid in the measuring tube is equal to the volume of luquid displaced from the wide vessel: (П8) nS1  h2 S 2 , where S1 is the area of the cross-section of the measuring tube, S 2 is the area of the cross-section of the wide vessel.

1 – vessel with working fluid, 2 – measuring tube along which the scale is located Fig. 3. Scheme of micromanometer with an inclined glass measuring tube From (A6), (A7) and (A8) we get:

 S  p  g h  g n sin   1   kg n , S 2   where k is a device constant;

139

(П9)

 S . k    sin   1  S 2  

(П10)

Values of k are given on the arc 3 (see Fig. 4, k = 0.2; 0.3; 0.4; 0.6; 0.8). The value of g  = 9.804 m/s2 corresponds to the geographical latitude of Almaty. Hence, the pressure p in Pa measured by the micromanometer with an inclined tube will be determined by the formula:

p  kg n

or

p  9,804k n .

(П11)

Here n is measured in mm. In the laboratory micromanometers of MMH type with variable inclination angle of the measuring tubes are used. Because of this the device has several measuring ranges (Fig. 4).

1 – a wide vessel, 2 – measuring tube, 3 – mounting arc, is used for fixing the tilt angle, 4 – base plate, 5 – scale of the measuring tube, 6 – level, 7 – screws for fixing the device by levels, 8 – displacer, used to set the fluid level against the zero scale mark Fig. 4. Scheme of micromanometer MMH The device consists of a wide vessel 1, measuring tube 2 fixed on the rotating arm, and attachment 3 for fixing the angle of inclination of the measuring tube. All parts are mounted on the common base 4. Each fixed angle of inclination of the measuring tube to the horizontal plane (i.e., the given measuring range) corresponds to a certain value of the device constant k, marked on the arc of the device 3.

140

141

4 5 6 7 8 9 10 11 12 13 14 15 100

3

2

Number of measure ments

Confidence probability

0.14 0.14 0.13 0.13 0.13 0.13 0.13 0.13 0.13 0.13 0.13 0.13 0.13 0.13

0.16

0.1

Values of Student’s coefficients 12.7 31.8 63.7 636.6

31.6 12.9 8.6 6.9 6.0 5.4 5.0 4.8 4.6 4.5 4.3 4.2 4.1 3.4

9.9 5.8 4.6 4.0 3.7 3.5 3.4 3.3 3.2 3.1 3.1 3.0 3.0 2.6

7 4.5 3.7 3.4 3.1 3.0 2.9 2.8 2.8 2.7 2.7 2.7 2.6 2.4

4.3 3.2 2.8 2.6 2.4 2.4 2.3 2.3 2.2 2.2 2.2 2.2 2.1 2.0

6.3 2.9 2.4 2.1 2.0 1.9 1.9 1.8 1.8 1.8 1.8 1.8 1.8 1.8 1.7

3.1 1.9 1.6 1.5 1.5 1.4 1.4 1.4 1.4 1.4 1.4 1.4 1.4 1.4 1.3

2.0 1.3 1.3 1.2 1.2 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.0

1.38 1.06

0.98 0.94 0.92 0.9 0.9 0.9 0.88 0.88 0.87 0.87 0.87 0.87 0.85

1

0.82 0.77 0.74 0.73 0.72 0.71 0.71 0.7 0.7 0.7 0.7 0.69 0.69 0.68

0.73 0.62 0.52 0.57 0.56 0.55 0.55 0.54 0.54 0.54 0.54 0.54 0.54 0.54 0.53

0.51 0.45 0.42 0.41 0.41 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.39 0.39 0.39

0.29 0.28 0.27 0.27 0.27 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.25

0.999

0.33

0.99

0.95

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.98

Student’s coefficients

APPENDIX 9

Еducational issue

Askarova Aliya Issatayev Sovet Bolegenova Saltanat Kashakarov Bladimir Korzun Irina Toleuov Gaziz Gumarova Sholpan Strautman Lydia Lavrichshev Oleg Issatayev Mukhtar Shortanbayeva Zhanar

MOLECULAR PHYSICS LABORATORY PRACTICUM IN PHYSICS Typesetting and cover design G. Кaliyeva

IB No.10242

Signed for publishing 08.12.2016. Format 60x84 1/12. Offset paper. Digital printing. Volume 8.87 printer’s sheet. 100 copies. Order No.5714. Publishing house «Qazaq university» Al-Farabi Kazakh National University KazNU, 71 Al-Farabi, 050040, Almaty Printed in the printing office of the «Kazakh University» publishing house

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